The equation of the plane passing through $(2,1,-1)$ is

$$a(x-2)+b(y-1)+c(z+1)=0\quad\text{.... (i)}$$

Since, this passes through $(-1,3,4)$.

$$\begin{array}{lr} \therefore & a(-1-2)+b(3-1)+c(4+1)=0 \\ \Rightarrow & -3 a+2 b+5 c=0\quad\text{.... (ii)} \end{array}$$

Since, the plane (i) is perpendicular to the plane $x-2 y+4 z=10$.

$$\begin{array}{lr} \therefore & 1 \cdot a-2 \cdot b+4 \cdot c=0 \\ \Rightarrow & a-2 b+4 c=0\quad\text{.... (iii)} \end{array}$$

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

$$\begin{aligned} \frac{a}{8+10} & =\frac{-b}{-17}=\frac{c}{4}=\lambda \\ \Rightarrow\quad a & =18 \lambda, b=17 \lambda, c=4 \lambda \end{aligned}$$

$$\begin{aligned} &\text { From Eq. (i), }\\ &\begin{aligned} & 18 \lambda(x-2)+17 \lambda(y-1)+4 \lambda(z+1) & =0 \\ \Rightarrow & 18 x-36+17 y-17+4 z+4 & =0 \\ \Rightarrow & 18 x+17 y+4 z-49 & =0 \\ \therefore & 18 x+17 y+4 z & =49 \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { We have, }\\ &\begin{aligned} \overrightarrow{\mathbf{r}} & =(8+3 \lambda) \hat{\mathbf{i}}-(9+16 \lambda) \hat{\mathbf{j}}+(10+7 \lambda) \hat{\mathbf{k}}) \\ & =8 \hat{\mathbf{i}}-9 \hat{\mathbf{j}}+10 \hat{\mathbf{k}}+3 \lambda \hat{\mathbf{i}}-16 \lambda \hat{\mathbf{j}}+7 \lambda \hat{\mathbf{k}} \\ & =8 \hat{\mathbf{i}}-9 \hat{\mathbf{j}}+10 \hat{\mathbf{k}}+\lambda(3 \hat{\mathbf{i}}-16 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}) \end{aligned} \end{aligned}$$

$$\begin{array}{ll} \Rightarrow & \overrightarrow{a_1}=8 \hat{\mathbf{i}}-9 \hat{\mathbf{j}}+10 \hat{\mathbf{k}} \text { and } \overrightarrow{\mathbf{b}_1}=3 \hat{\mathbf{i}}-16 \hat{\mathbf{j}}+7 \hat{\mathbf{k}} \quad\text{.... (i)}\\ \text { Also } & \overrightarrow{\mathbf{r}}=15 \hat{\mathbf{i}}+29 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}+\mu(3 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}) \\ \Rightarrow & \overrightarrow{\mathbf{a}_2}=15 \hat{\mathbf{i}}+29 \hat{\mathbf{j}}+5 \hat{\mathbf{k}} \text { and } \overrightarrow{\mathbf{b}_2}=3 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}\quad\text{.... (ii)} \end{array}$$

Now, shortest distance betwen two lines is given by $\left|\frac{\left(\overrightarrow{\mathbf{b}_1} \times \overrightarrow{\mathbf{b}_2}\right) \cdot\left(\overrightarrow{\mathbf{a}_2}-\overrightarrow{\mathbf{a}_1}\right)}{\left|\overrightarrow{\mathbf{b}_1} \times \overrightarrow{\mathbf{b}_2}\right|}\right|$

$$\begin{aligned} \therefore\quad \overrightarrow{\mathbf{b}_1} \times \overrightarrow{\mathbf{b}_2} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{array}\right| \\ & =\hat{\mathbf{i}}(80-56)-\hat{\mathbf{j}}(-15-21)+\hat{\mathbf{k}}(24+48) \\ & =24 \hat{\mathbf{i}}+36 \hat{\mathbf{j}}+72 \hat{\mathbf{k}} \end{aligned}$$

Now,

$$\begin{aligned} \left|\overrightarrow{\mathbf{b}_1} \times \overrightarrow{\mathbf{b}_2}\right| & =\sqrt{(24)^2+(36)^2+(72)^2} \\ & =12 \sqrt{2^2+3^2+6^2}=84 \end{aligned}$$

and

$$\begin{aligned} \left(\overrightarrow{\mathbf{a}_2}-\overrightarrow{\mathbf{a}_1}\right) & =(15-8) \hat{\mathbf{i}}+(29+9) \hat{\mathbf{j}}+(5-10) \hat{\mathbf{k}} \\ & =7 \hat{\mathbf{i}}+38 \hat{\mathbf{j}}-5 \hat{\mathbf{k}} \end{aligned}$$

$$\begin{aligned} \therefore \quad \text { Shortest distance } & =\left|\frac{(24 \hat{\mathbf{i}}+36 \hat{\mathbf{j}}+72 \hat{\mathbf{k}}) \cdot(7 \hat{\mathbf{i}}+38 \hat{\mathbf{j}}-5 \hat{\mathbf{k}})}{84}\right| \\ & =\left|\frac{168+1368-360}{84}\right|=\left|\frac{1176}{84}\right|=14 \text { units } \end{aligned}$$

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.

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

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