$$\begin{aligned} &\text { Eliminating } n \text { from both the equations, we have }\\ &\begin{aligned} & l^2+m^2-(l-m)^2=0 \\ & \Rightarrow \quad l^2+m^2-l^2-m^2+2 m l=0 \quad \Rightarrow \quad 2 l m=0 \end{aligned} \end{aligned}$$

$$ \begin{aligned} & \Rightarrow \quad l m=0 \quad \Rightarrow \quad(-m-n) m=0 \quad[\because l=-m-n] \\ & \Rightarrow \quad(m+n) m=0 \\ & \Rightarrow \quad m=-n \Rightarrow m=0 \\ & \Rightarrow \quad l=0, l=-n \end{aligned}$$

Thus, Dr's two lines are proportional to $0,-n, n$ and $-n, 0, n$ i.e., $0,-1,1$ and $-1,0,1$.

So, the vector parallel to these given lines are $\overrightarrow{\mathbf{a}}=-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}}=-\hat{\mathbf{i}}+\hat{\mathbf{k}}$

$$ \begin{aligned} &\text { Now, }\\ &\begin{aligned} \cos \theta & =\frac{\vec{a} \vec{b}}{|\vec{a}||\vec{b}|}=\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \Rightarrow \cos \theta=\frac{1}{2} \\ \therefore\quad \theta & =\frac{\pi}{3} \quad\left[\because \cos \frac{\pi}{3}=\frac{1}{2}\right] \end{aligned} \end{aligned}$$

We have $l, m, n$ and $l+\delta l, m+\delta m, n+\delta n$ as direction cosines of a variable line in two different positions.

$$\begin{array}{lr} \therefore & l^2+m^2+n^2=1 \quad\text{.... (i)}\\ \text { and } & (l+\delta l)^2+(m+\delta m)^2+(n+\delta n)^2=1 \quad\text{.... (ii)} \end{array}$$

$$\begin{array}{lcc} \Rightarrow & l^2+m^2+n^2+\delta l^2+\delta m^2+\delta n^2+2(l \delta l+m \delta m+n \delta n)=1 \\ \Rightarrow & \delta l^2+\delta m^2+\delta n^2=-2(l \delta l+m \delta m+n \delta n) & \quad\left[\because l^2+m^2+n^2=1\right] \\ \Rightarrow & l \delta l+m \delta m+n \delta n=\frac{-1}{2}\left(\delta l^2+\delta m^2+\delta n^2\right) & \ldots(\text { iii }) \end{array}$$

Now, $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathrm{b}}$ are unit vectors along a line with direction cosines $l, m, n$ and $(l+\delta l),(m+\delta m),(n+\delta n)$, respectively.

$$\begin{array}{ll} \therefore & \overrightarrow{\mathbf{a}}=l \hat{\mathbf{i}}+m \hat{\mathbf{j}}+n \hat{\mathbf{k}} \text { and } \overrightarrow{\mathbf{b}}=(l+\delta l) \hat{\mathbf{i}}+(m+\delta m) \hat{\mathbf{j}}+(n+\delta n) \hat{\mathbf{k}} \\ \Rightarrow & \cos \delta \theta=\frac{\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}|}=\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} \quad[\because|\hat{\mathbf{a}}|=|\hat{\mathbf{b}}|=1] \end{array}$$

$$ \begin{aligned} &\begin{aligned} \Rightarrow \quad \cos \delta \theta & =l(l+\delta l)+m(m+\delta m)+n(n+\delta n) \\ & =\left(l^2+m^2+n^2\right)+(l \delta l+m \delta m+n \delta n) \\ & =1-\frac{1}{2}\left(\delta l^2+\delta m^2+\delta n^2\right) \text { [using Eq. (iii)] } \end{aligned}\\ \end{aligned}$$

$$\begin{array}{lcc} \Rightarrow & 2(1-\cos \delta \theta)=\left(\delta l^2+\delta m^2+\delta n^2\right) \\ \Rightarrow & 2 \cdot 2 \sin ^2 \frac{\delta \theta}{2}=\delta l^2+\delta m^2+\delta n^2 & \quad\left[\because 1-\cos \theta=2 \sin ^2 \frac{\theta}{2}\right] \\ \Rightarrow & 4\left(\frac{\delta \theta}{2}\right)^2=\delta l^2+\delta m^2+\delta n^2 & {\left[\text { since, } \frac{\delta \theta}{2} \text { is small, then } \sin \frac{\delta \theta}{2}=\frac{\delta \theta}{2}\right]}\\ \therefore & \delta \theta^2=\delta l^2+\delta m^2+\delta n^2 \end{array}$$

Since, DC's of line OA are $\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}$ and $\frac{c}{\sqrt{a^2+b^2+c^2}}$.

Also, $$\overrightarrow{\mathbf{n}}=\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathbf{a}}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}$$

The equation of plane passes through $(a, b, c)$ and perpendicular to $O A$ is given by

$$\begin{array}{rlrl} & {[\overrightarrow{\mathbf{r}}-\overrightarrow{\mathbf{a}}] \cdot \vec{n}} =0 \\ \Rightarrow & \overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}} =\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{n}} \\ \Rightarrow & {[(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot(a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}})]} =(a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}) \cdot(a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}) \\ \Rightarrow & a x+b y+c z =a^2+b^2+c^2 \end{array}$$

Consider OX, OY, OZ and ox, oy, oz are two system of rectangular axes.

Let their corresponding equation of plane be

$$\begin{aligned} & \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 \quad\text{.... (i)}\\ \text{and}\quad & \frac{x}{a^{\prime}}+\frac{y}{b^{\prime}}+\frac{z}{c^{\prime}}=1\quad\text{.... (ii)} \end{aligned}$$

Also, the length of perpendicular from origin to Eqs. (i) and (ii) must be same.

$$\therefore \quad \frac{\frac{0}{a}+\frac{0}{b}+\frac{0}{c}-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}=\frac{\frac{0}{a^{\prime}}+\frac{0}{b^{\prime}}+\frac{0}{c^{\prime}}-1}{\sqrt{\frac{1}{a^{\prime 2}}+\frac{1}{b^{\prime 2}}+\frac{1}{c^{\prime 2}}}}$$

$$\begin{array}{ll} \Rightarrow & \sqrt{\frac{1}{a^{2^2}}+\frac{1}{b^{\prime 2}}+\frac{1}{c^{\prime^2}}}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}} \\ \Rightarrow & \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{a^{\prime^2}}+\frac{1}{b^{2^2}}+\frac{1}{c^{\prime 2}} \end{array}$$

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,

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