We have, equation of line as $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$

$$\begin{array}{l} \Rightarrow & \frac{x-4}{-2} & =\frac{y}{6}=\frac{z-1}{-3}=\lambda \\ \Rightarrow & x & =-2 \lambda+4, y=6 \lambda \text { and } z=-3 \lambda+1 \end{array}$$

Let the coordinates of $L$ be $(4-2 \lambda, 6 \lambda, 1-3 \lambda)$ and direction ratios of $P L$ are proportional to $$(4-2 \lambda-2,6 \lambda-3,1-3 \lambda+8) \text { i.e., }(2-2 \lambda, 6 \lambda-3,9-3 \lambda).$$

Also, direction ratios are proportional to $-2,6,-3$. Since, $P L$ is perpendicular to give line.

$$\begin{array}{llrl} \therefore & -2(2-2 \lambda)+6(6 \lambda-3)-3(9-3 \lambda) =0 \\ \Rightarrow & -4+4 \lambda+36 \lambda-18-27+9 \lambda =0 \\ \Rightarrow & 49 \lambda =49 \Rightarrow \lambda=1 \end{array}$$

So, the coordinates of $L$ are $(4-2 \lambda, 6 \lambda, 1-3 \lambda)$ i.e., $(2,6,-2)$.

Also,

$$\begin{aligned} \text { length of } P L & =\sqrt{(2-2)^2+(6-3)^2+(-2+8)^2} \\ & =\sqrt{0+9+36}=3 \sqrt{5} \text { units } \end{aligned}$$

We have, equation of the line as $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}=\lambda$

$$\Rightarrow \quad x=\lambda-5, y=4 \lambda-3, z=6-9 \lambda$$

Let the coordinates of $L$ be $(\lambda-5,4 \lambda-3,6-9 \lambda)$, then Dr's of $P L$ are $(\lambda-7,4 \lambda-7,7-9 \lambda)$.

Also, the direction ratios of given line are proportional to $1, 4, -9$.

Since, $P L$ is perpendicular to the given line.

$$\begin{array}{rlrl} \therefore & (\lambda-7) \cdot 1+(4 \lambda-7) \cdot 4+(7-9 \lambda) \cdot(-9) =0 \\ \Rightarrow & \lambda-7+16 \lambda-28+81 \lambda-63 =0 \\ \Rightarrow & 98 \lambda =98 \Rightarrow \lambda=1 \end{array}$$

$$\begin{aligned} &\text { So, the coordinates of } L \text { are }(-4,1,-3) \text {. }\\ &\begin{aligned} \therefore \quad \text { Required distance, } P L & =\sqrt{(-4-2)^2+(1-4)^2+(-3+1)^2} \\ & =\sqrt{36+9+4}=7 \text { units } \end{aligned} \end{aligned}$$

Equation of the given plane is $2 x-2 y+4 z+5=0\quad\text{.... (i)}$

$$\Rightarrow \quad \vec{n}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$$

So, the equation of line through $\left(1, \frac{3}{2}, 2\right)$ and parallel to $\overrightarrow{\mathbf{n}}$ is given by

$$\begin{aligned} \frac{x-1}{2} & =\frac{y-3 / 2}{-2}=\frac{z-2}{4}=\lambda \\ \Rightarrow\quad x & =2 \lambda+1, y=-2 \lambda+\frac{3}{2} \text { and } z=4 \lambda+2 \end{aligned}$$

$$ \begin{aligned} &\text { If this point lies on the given plane, then }\\ &\begin{aligned} & 2(2 \lambda+1)-2\left(-2 \lambda+\frac{3}{2}\right)+4(4 \lambda+2)+5 =0 \quad\text{[using Eq. (i)]}\\ \Rightarrow \quad & 4 \lambda+2+4 \lambda-3+16 \lambda+8+5 =0 \\ \Rightarrow \quad & 24 \lambda=-12 \Rightarrow \lambda=\frac{-1}{2} \end{aligned} \end{aligned}$$

$$\begin{aligned} &\therefore \quad \text { Required foot of perpendicular }\\ &=\left[2 \times\left(\frac{-1}{2}\right)+1,-2 \times\left(\frac{-1}{2}\right)+\frac{3}{2}, 4 \times\left(\frac{-1}{2}\right)+2\right] \text { i.e., }\left(0, \frac{5}{2}, 0\right)\\ &\begin{aligned} \therefore \quad \text { Required length of perpendicular } & =\sqrt{(1-0)^2+\left(\frac{3}{2}-\frac{5}{2}\right)^2+(2-0)^2} \\ & =\sqrt{1+1+4}=\sqrt{6} \text { units } \end{aligned} \end{aligned}$$

Equation of the two planes are $x+2 y=0$ and $3 y-z=0$.

Let $\overrightarrow{\mathbf{n}_1}$ and $\overrightarrow{\mathbf{n}_2}$ are the normals to the two planes, respectively.

$$\therefore \quad \overrightarrow{\mathbf{n}_1}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}} \text { and } \overrightarrow{\mathbf{n}_2}=3 \hat{\mathbf{j}}-\hat{\mathbf{k}}$$

Since, required line is parallel to the given two planes.

$$\begin{aligned} &\text { Therefore, }\\ &\begin{aligned} \overrightarrow{\mathbf{b}} & =\overrightarrow{\mathbf{n}_1} \times \overrightarrow{\mathbf{n}_2}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & 0 \\ 0 & 3 & -1 \end{array}\right| \\ & =\hat{\mathbf{i}}(-2)-\hat{\mathbf{j}}(-1)+\hat{\mathbf{k}}(3) \\ & =-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}} \end{aligned} \end{aligned}$$

$$\begin{aligned} &\text { So, the equation of the lines through the point }(3,0,1) \text { and parallel to the given two planes are }\\ &\begin{array}{ll} & (x-3) \hat{\mathbf{i}}+(y-0) \hat{\mathbf{j}}+(z-1) \hat{\mathbf{k}}+\lambda(-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \\ \Rightarrow \quad & (x-3) \hat{\mathbf{i}}+y \hat{\mathbf{j}}+(z-1) \hat{\mathbf{k}}+\lambda(-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \end{array} \end{aligned}$$

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

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

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