The equation of a plane through the line of intersection of the planes $x+2 y+3 z-4=0$ and $2 x+y-z+5=0$ is

$$\begin{aligned} & (x+2 y+3 z-4)+\lambda(2 x+y-z+5) =0\\ \Rightarrow & x(1+2 \lambda)+y(2+\lambda)+z(-\lambda+3)-4+5 \lambda=0\quad\text{... (i)} \end{aligned}$$

$$\begin{aligned} &\text { Also, this is perpendicular to the plane } 5 x+3 y+6 z+8=0 \text {. }\\ &\begin{array}{lll} \therefore & 5(1+2 \lambda)+3(2+\lambda)+6(3-\lambda)=0 & {\left[\because a_1 a_2+b_1 b_2+c_1 c_2=0\right]} \\ \Rightarrow & 5+10 \lambda+6+3 \lambda+18-6 \lambda=0 & \end{array} \end{aligned}$$

$$\therefore \quad \lambda=-29 / 7$$

From Eq. (i),

$$\begin{array}{cc} & x\left[1+2\left(\frac{-29}{7}\right)\right]+y\left(2-\frac{29}{7}\right)+z\left(\frac{29}{7}+3\right)-4+5\left(\frac{-29}{7}\right)=0 \\ \Rightarrow & x(7-58)+y(14-29)+z(29+21)-28-145=0 \\ \Rightarrow & -51 x-15 y+50 z-173=0 \end{array}$$

So, the required equation of plane is $51 x+15 y-50 z+173=0$.

Equation of the plane is $a x+b y=0\quad\text{.... (i)}$

$\therefore$ Equation of the plane after new position is

$$\frac{a x \cos \alpha}{\sqrt{a^2+b^2}}+\frac{b y \cos \alpha}{\sqrt{b^2+a^2}} \pm z \sin \alpha=0$$

$$\begin{array}{llr} \Rightarrow & \frac{a x}{\sqrt{a^2+b^2}}+\frac{b y}{\sqrt{b^2+a^2}} \pm z \tan \alpha=0 & \text { [on dividing by } \cos \alpha \text { ] } \\ \Rightarrow & a x+b y \pm z \tan \alpha \sqrt{\alpha^2+b^2}=0 & \text { [on multiplying with } \sqrt{a^2+b^2} \text { ] } \end{array}$$

Alternate Method

$$\begin{aligned} \text{Given, planes are}\quad a x+b y & =0 \quad\text{.... (i)}\\ \text{and}\quad z & =0\quad\text{.... (ii)} \end{aligned}$$

Therefore, the equation of any plane passing through the line of intersection of planes (i) and (ii) may be taken as $a x+b y+k=0\quad\text{.... (iii)}$.

Then, direction cosines of a normal to the plane (iii) are $\frac{a}{\sqrt{a^2+b^2+k^2}}, \frac{b}{\sqrt{a^2+b^2+k^2}}$, $\frac{c}{\sqrt{a^2+b^2+k^2}}$ and direction cosines of the normal to the plane (i) are $\frac{a}{\sqrt{a^2+b^2}}, \frac{b}{\sqrt{a^2+b^2}}, 0$.

Since, the angle between the planes (i) and (ii) is $\alpha$,

$$\begin{aligned} &\begin{aligned} \therefore \quad \cos \alpha & =\frac{a \cdot a+b \cdot b+k \cdot 0}{\sqrt{a^2+b^2+k^2} \sqrt{a^2+b^2}} \\ & =\sqrt{\frac{a^2+b^2}{a^2+b^2+k^2}} \\ \Rightarrow \quad k^2 \cos ^2 \alpha & =a^2\left(1-\cos ^2 \alpha\right)+b^2\left(1-\cos ^2 \alpha\right) \\ \Rightarrow \quad k^2 & =\frac{\left(a^2+b^2\right) \sin ^2 \alpha}{\cos ^2 \alpha} \\ k & = \pm \sqrt{a^2+b^2} \tan \alpha \end{aligned}\\ &\text { On putting this value in plane (iii), we get the equation of the plane as }\\ &a x+b y+z \sqrt{a^2+b^2} \tan \alpha=0 \end{aligned}$$

We have, $\quad \overrightarrow{\mathbf{n}_1}=(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}), d_1=6$ and $\overrightarrow{\mathbf{n}_2}=(3 \hat{\mathbf{i}}-\hat{\mathbf{j}}-4 \hat{\mathbf{k}}), d_2=0$

Using the relation, $\quad \overrightarrow{\mathbf{r}} \cdot\left(\overrightarrow{\mathbf{n}}_1+\lambda \overrightarrow{\mathbf{n}_2}\right)=d_1+d_2 \lambda$ $$ \begin{array}{ll} \Rightarrow & \overrightarrow{\mathbf{r}} \cdot[(\hat{\mathbf{i}}+3 \hat{\mathbf{j}})+\lambda(3 \hat{\mathbf{i}}-\hat{\mathbf{j}}-4 \hat{\mathbf{k}})]=6+0 \cdot \lambda \\ \Rightarrow & \overrightarrow{\mathbf{r}} \cdot[(1+3 \lambda) \hat{\mathbf{i}}+(3-\lambda) \hat{\mathbf{j}}+\hat{\mathbf{k}}(-4 \lambda)]=6\quad\text{.... (i)}] \end{array}$$

On dviding both sides by $\sqrt{(1+3 \lambda)^2+(3-\lambda)^2+(-4 \lambda)^2}$, we get

$$\frac{\overrightarrow{\mathbf{r}} \cdot[(1+3 \lambda) \hat{\mathbf{i}}+(3-\lambda) \hat{\mathbf{j}}+\hat{\mathbf{k}}(-4 \lambda)]}{\sqrt{(1+3 \lambda)^2+(3-\lambda)^2+(-4 \lambda)^2}}=\frac{6}{\sqrt{(1+3 \lambda)^2+(3-\lambda)^2+(-4 \lambda)^2}}$$

Since, the perpendicular distance from origin is unity.

$$\begin{array}{lrl} \therefore & \frac{6}{\sqrt{(1+3 \lambda)^2+(3-\lambda)^2+(-4 \lambda)^2}}=1 \\ \Rightarrow & (1+3 \lambda)^2+(3-\lambda)^2+(-4 \lambda)^2 & =36 \\ \Rightarrow & 1+9 \lambda^2+6 \lambda+9+\lambda^2-6 \lambda+16 \lambda^2=36 \\ \Rightarrow & 26 \lambda^2+10=36 \\ \Rightarrow & \lambda^2=1 \\ \therefore & \lambda & = \pm 1 \end{array}$$

$$\begin{aligned} &\text { Using Eq. (i), the required equation of plane is }\\ &\begin{array}{lrl} & \overrightarrow{\mathbf{r}} \cdot[(1 \pm 3) \hat{\mathbf{i}}+(3 \mp 1) \hat{\mathbf{j}}+(\mp 4) \hat{\mathbf{k}}] & =6 \\ \Rightarrow & \overrightarrow{\mathbf{r}} \cdot[(1+3) \hat{\mathbf{i}}+(3-1) \hat{\mathbf{j}}+(-4) \hat{\mathbf{k}}] & =6 \\ \text { and } & \overrightarrow{\mathbf{r}} \cdot[(1-3) \hat{\mathbf{i}}+(3+1) \hat{\mathbf{j}}+4 \hat{\mathbf{k}}] & =6 \\ \Rightarrow & \overrightarrow{\mathbf{r}} \cdot(4 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}) & =6 \\ \text { and } & \overrightarrow{\mathbf{r}} \cdot(-2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) & =6 \\ \Rightarrow & 4 x+2 y-4 z-6 & =0 \\ \text { and } & -2 x+4 y+4 z-6 & =0 \end{array} \end{aligned}$$

To show that these given points $(\hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}})$ and $3(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})$ are equidistant from the plane $\overrightarrow{\mathbf{r}} \cdot(5 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-7 \hat{\mathbf{k}})+9=0$, we first find out the mid- point of the points which is $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$.

On substituting $\vec{r}$ by the mid-point in plane, we get

$$\begin{aligned} \text { LHS } & =(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \cdot(5 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-7 \hat{\mathbf{k}})+9 \\ & =10+2-21+9=0 \\ & =\text { RHS } \end{aligned}$$

Hence, the two points lie on opposite sides of the plane are equidistant from the plane.

We have, $$\overrightarrow{A B}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}} \text { and } \overrightarrow{C D}=-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$$

Also, the position vectors of $A$ and $C$ are $6 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ and $-9 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$, respectively. Since, $\overrightarrow{\mathrm{PQ}}$ is perpendicular to both $\overrightarrow{A B}$ and $\overrightarrow{C D}$.

So, $P$ and $Q$ will be foot of perpendicular to both the lines through $A$ and $C$.

Now, equation of the line through $A$ and parallel to the vector $\overrightarrow{A B}$ is,

$$\overrightarrow{\mathbf{r}}=(6 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})+\lambda(3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})$$

$$\begin{aligned} &\text { and the line through } C \text { and parallel to the vector } \overrightarrow{C D} \text { is given by }\\ &\begin{array}{ll} & \overrightarrow{\mathbf{r}}=-9 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}+\mu(-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \quad\text{.... (i)}]\\ \text { Let } & \overrightarrow{\mathbf{r}}=(6 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})+\lambda(3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\ \text { and } & \overrightarrow{\mathbf{r}}=-9 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}+\mu(-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})\quad\text{..... (ii)} \end{array} \end{aligned}$$

Let $P(6+3 \lambda, 7-\lambda, 4+\lambda)$ is any point on the first line and $Q$ be any point on second line is given by $(-3 \mu,-9+2 \mu, 2+4 \mu)$.

$$\begin{aligned} \therefore \quad \overrightarrow{\mathrm{PQ}} & =(-3 \mu-6-3 \lambda) \hat{\mathbf{i}}+(-9+2 \mu-7+\lambda) \hat{\mathbf{j}}+(2+4 \mu-4-\lambda) \hat{\mathbf{k}} \\ & =(-3 \mu-6-3 \lambda) \hat{\mathbf{i}}+(2 \mu+\lambda-16) \hat{\mathbf{j}}+(4 \mu-\lambda-2) \hat{\mathbf{k}} \end{aligned}$$

If $\overrightarrow{P Q}$ is perpendicular to the first line, then

$$\begin{aligned} &\begin{array}{lr} & 3(-3 \mu-6-3 \lambda)-(2 \mu+\lambda-16)+(4 \mu-\lambda-2)=0 \\ \Rightarrow & -9 \mu-18-9 \lambda \square-2 \mu-\lambda+16+4 \mu-\lambda-2=0 \\ \Rightarrow & -7 \mu-11 \lambda-4=0\quad\text{.... (iii)} \end{array}\\ &\text { If } \overrightarrow{P Q} \text { is perpendicular to the second line, then }\\ &\begin{aligned} & -3(-3 \mu-6-3 \lambda)+2(2 \mu+\lambda-16)+4(4 \mu-\lambda-2) =0 \\ \Rightarrow & 9 \mu+18+9 \lambda+4 \mu+2 \lambda-32+16 \mu-4 \lambda-8 =0 \\ \Rightarrow & 29 \mu+7 \lambda-22 =0\quad\text{.... (iv)} \end{aligned}\end{aligned}$$

On solving Eqs. (iii) and (iv), we get

$$\begin{array}{rrr} & -49 \mu-77 \lambda-28 & =0 \\ \Rightarrow & 319 \mu+77 \lambda-242 & =0 \\ \Rightarrow & 270 \mu-270 & =0 \\ \Rightarrow & \mu & =1 \end{array}$$

Using $\mu$ in Eq. (iii), we get

$$\begin{array}{r} & -7(1)-11 \lambda-4 =0 \\ \Rightarrow & -7-11 \lambda-4 =0 \\ \Rightarrow & -11-11 \lambda =0 \\ \Rightarrow & \lambda =-1 \end{array}$$

$$\begin{aligned} \therefore \quad \overrightarrow{P Q} & =[-3(1)-6-3(-1)] \hat{\mathbf{i}}+[2(1)+(-1)-16] \hat{\mathbf{j}}+[4(1)-(-1)-2] \hat{\mathbf{k}} \\ & =-6 \hat{\mathbf{i}}-15 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \end{aligned}$$