We know that, the cartesian equation of a line that passes through two points ( $x_1, y_1, z_1$ ) and $\left(x_2, y_2, z_2\right)$ is

$$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$$

Hence, the cartesian equation of line passes through $A(0,-1,-1)$ and $B(4,5,1)$ is

$$\begin{aligned} \frac{x-0}{4-0} & =\frac{y+1}{5+1}=\frac{z+1}{1+1} \\ \Rightarrow\quad \frac{x}{4} & =\frac{y+1}{6}=\frac{z+1}{2}\quad\text{.... (i)} \end{aligned}$$

and cartesian equation of the line passes through $C(3,9,4)$ and $D(-4,4,4)$ is

$$\begin{aligned} &\frac{x-3}{-4-3}=\frac{y-9}{4-9}=\frac{z-4}{4-4} \\ \Rightarrow\quad & \frac{x-3}{-7}=\frac{y-9}{-5}=\frac{z-4}{0}\quad\text{.... (ii)} \end{aligned}$$

If the lines intersect, then shortest distance between both of them should be zero.

$\therefore$ Shortest distance between the lines

$=\frac{\left|\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2\end{array}\right|}{\sqrt{\left(b_1 c_2-b_2 c_1\right)^2+\left(c_1 a_2-c_2 a_1\right)^2+\left(a_1 b_2-a_2 b_1\right)^2}}$

$$\begin{aligned} & =\frac{\left|\begin{array}{ccc} 3-0 & 9+1 & 4+1 \\ 4 & 6 & 2 \\ -7 & -5 & 0 \end{array}\right|}{\sqrt{(6 \cdot 0+10)^2+(-14-0)^2+(-20+42)^2}} \\ & =\frac{\left|\begin{array}{ccc} 3 & 10 & 5 \\ 4 & 6 & 2 \\ -7 & -5 & 0 \end{array}\right|}{\sqrt{100+196+484}} \\ & =\frac{3(0+10)-10(14)+5(-20+42)}{\sqrt{780}} \\ & =\frac{30-140+110}{\sqrt{780}=0} \end{aligned}$$

So, the given lines intersect.

$$\begin{aligned} \text{We have,}\quad & x=p y+q \Rightarrow y=\frac{x-q}{p} \quad\text{.... (i)}\\ \text{and}\quad & z=r y+s \Rightarrow y=\frac{z-s}{r}\quad\text{.... (ii)} \end{aligned}$$

$$\begin{array}{ll} \Rightarrow & \frac{x-q}{p}=\frac{y}{1}=\frac{z-s}{r} \quad \text { [using Eqs. (i) and (ii)] ...(iii) }\\ \text { Similarly, } & \frac{x-q^{\prime}}{p^{\prime}}=\frac{y}{1}=\frac{z-s^{\prime}}{r^{\prime}}\quad\text{.... (iv)} \end{array}$$

From Eqs. (iii) and (iv),

$$\begin{aligned} & a_1=p, b_1=1, c_1=r \\ \text{and}\quad & a_2=p^{\prime}, b_2=1, c_2=r^{\prime} \end{aligned}$$

If these given lines are perpendicular to each other, then

$$\begin{aligned} a_1 a_2+b_1 b_2+c_1 c_2 & =0 \\ \Rightarrow\quad p p^{\prime}+1+r r^{\prime} & =0 \end{aligned}$$

which is the required condition.

Since, the equation of a plane is bisecting perpendicular the line joining the points $A(2,3,4)$ and $B(4,5,8)$ at right angles.

So, mid-point of $A B$ is $\left(\frac{2+4}{2}, \frac{3+5}{2}, \frac{4+8}{2}\right)$ i.e., $(3,4,6)$.

Also, $$\overrightarrow{\mathbf{N}}=(4-2) \hat{\mathbf{i}}+(5-3) \hat{\mathbf{j}}+(8-4) \hat{\mathbf{k}}=2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$$

So, the required equation of the plane is $(\overrightarrow{\mathbf{r}}-\overrightarrow{\mathbf{a}}) \cdot \overrightarrow{\mathbf{N}}=0$.

$$\begin{array}{lrrr} \Rightarrow & {[(x-3) \hat{\mathbf{i}}+(y-4) \hat{\mathbf{j}}+(z-6) \hat{\mathbf{k}}] \cdot(2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})} & =0 & {[\because \overrightarrow{\mathbf{a}}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}]} \\ \Rightarrow & 2 x-6+2 y-8+4 z-24 & =0 \\ \Rightarrow & 2 x+2 y+4 z=38 \\ \therefore & x+y+2 z=19 \end{array}$$

Since, normal to the plane is equally inclined to the coordinate axis.

Therefore, $$\cos \alpha=\cos \beta=\cos \gamma=\frac{1}{\sqrt{3}}$$

So, the normal is $\overrightarrow{\mathbf{N}}=\frac{1}{\sqrt{3}} \hat{\mathbf{i}}+\frac{1}{\sqrt{3}} \hat{\mathbf{j}}+\frac{1}{\sqrt{3}} \hat{\mathbf{k}}$ and plane is at a distance of $3 \sqrt{3}$ units from origin.

The equation of plane is $$\overrightarrow{\mathbf{r}} \cdot \hat{\mathbf{N}}=3 \sqrt{3}\quad \left[\because \hat{\mathbf{N}}=\frac{\overrightarrow{\mathbf{N}}}{|\mathbf{N}|}\right]$$

[since, vector equation of the plane at a distance $p$ from the origin is $\overrightarrow{\mathbf{r}} \cdot \hat{\mathbf{N}}=p$]

$$\begin{array}{rlrl} \Rightarrow & (x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot \frac{\left(\frac{1}{\sqrt{3}} \hat{\mathbf{i}}+\frac{1}{\sqrt{3}} \hat{\mathbf{j}}+\frac{1}{\sqrt{3}} \hat{\mathbf{k}}\right)}{1} =3 \sqrt{3} \\ \Rightarrow & \frac{x}{\sqrt{3}}+\frac{y}{\sqrt{3}}+\frac{z}{\sqrt{3}} =3 \sqrt{3} \\ \therefore & x+y+z =3 \sqrt{3} \cdot \sqrt{3}=9 \end{array}$$

So, the required equation of plane is $x+y+z=9$.

Since, the line drawn from the point $(-2,-1,-3)$ meets a plane at right angle at the point $(1,-3,3)$. So, the plane passes through the point $(1,-3,3)$ and normal to plane is $(-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-6 \hat{\mathbf{k}})$.

$$\begin{aligned} &\begin{array}{ll} \Rightarrow & \overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\ \text { and } & \overrightarrow{\mathbf{N}}=-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-6 \hat{\mathbf{k}} \end{array}\\ &\text { So, the equation of required plane is }(\vec{r}-\overrightarrow{\mathbf{a}}) \cdot \overrightarrow{\mathbf{N}}=0\\ &\begin{aligned} \Rightarrow & {[(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}})-(\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})] \cdot(-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}) } =0 \\ \Rightarrow & {[(x-1) \hat{\mathbf{i}}+(y+3) \hat{\mathbf{j}}+(z-3) \hat{\mathbf{k}}] \cdot(-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}) } =0 \\ \Rightarrow & -3 x+3+2 y+6-6 z+18 =0 \\ \Rightarrow & -3 x+2 y-6 z =-27 \\ \therefore & -2 y+6 z-27 =0 \end{aligned} \end{aligned}$$