(i)

Normal of $\pi_2=\left(\begin{array}{l}0 \\ 1 \\ 1\end{array}\right) \times\left(\begin{array}{c}1 \\ -1 \\ -1\end{array}\right)=\left(\begin{array}{c}0 \\ 1 \\ -1\end{array}\right)$ or $\left(\begin{array}{c}0 \\ 1 \\ -1\end{array}\right)$ using G.C.

$\pi_2: \mathbf{r} \cdot\left(\begin{array}{c}0 \\ 1 \\ -1\end{array}\right)=\left(\begin{array}{l}2 \\ 1 \\ 0\end{array}\right) \cdot\left(\begin{array}{c}0 \\ 1 \\ -1\end{array}\right)=1$

Using G.C. to solve
$4 x+3 y+5 z=7$
$y-z=1$

We obtain

$l_1: \mathbf{r}=\left(\begin{array}{l}1 \\ 1 \\ 0\end{array}\right)+\lambda\left(\begin{array}{c}-2 \\ 1 \\ 1\end{array}\right), \lambda \in \mathbb{R}$

$\left(\begin{array}{c}-1 \\ 2 \\ 1\end{array}\right)=\left(\begin{array}{l}1 \\ 1 \\ 0\end{array}\right)+\lambda\left(\begin{array}{c}-2 \\ 1\end{array}\right)$

Since $\lambda=1$ satisfy the above equation, $P(-1,2,1)$ lies on line $l_1$

(ii)

Direction vector of $l_2=\underbrace{\left(\begin{array}{c}0 \\ 1 \\ -1\end{array}\right)}_{\text {normal to } \pi_2} \times \underbrace{\left(\begin{array}{c}-2 \\ 1 \\ 1\end{array}\right)}_{\text {direction vector of } l_1}=\left(\begin{array}{l}2 \\ 2 \\ 2\end{array}\right)=2\left(\begin{array}{l}1 \\ 1 \\ 1\end{array}\right)$

$\mathbf{r}=\left(\begin{array}{c}-1 \\ 2 \\ 1\end{array}\right)+\mu\left(\begin{array}{l}1 \\ 1 \\ 1\end{array}\right), \mu \in \mathbb{R}$
$x+1=y-2=z-1$

Position vector of any point on $l_2$ takes the form $\left(\begin{array}{c}-1+\mu \\ 2+\mu \\ 1+\mu\end{array}\right)$ Distance $(-1+\mu, 2+\mu, 1+\mu)$ from $\pi_1$

$=\frac{\mid\left( \begin{array}{c}-1+\mu \\ 2+\mu \\ 1+\mu\end{array}\right) \cdot\left(\begin{array}{l}4 \\ 3 \\ 5\end{array}\right)-7 \mid}{\sqrt{50}}$

$=\frac{|4(-1+\mu)+3(2+\mu)+5(1+\mu)-7|}{\sqrt{50})}=\frac{12|\mu|}{\sqrt{50}}$

When $\frac{12|\mu|}{\sqrt{50}}=3 \sqrt{2}$
$|\mu|=\frac{30}{12}=\frac{5}{2}$
Thus $\mu=\frac{5}{2}$ or $-\frac{5}{2}$
Points on $l_2$ are $\left(\frac{3}{2}, \frac{9}{2}, \frac{7}{2}\right),\left(-\frac{7}{2},-\frac{1}{2},-\frac{3}{2}\right)$

Since $a, b$ and $c$ are negative, thus $Q\left(-\frac{7}{2},-\frac{1}{2},-\frac{3}{2}\right)$ Let $x$ be the length of projection of $\overrightarrow{O P}$ on $\pi_1$

Using Pythagoras Theorem
$(|\overrightarrow{Q P}|)^2=(3 \sqrt{2})^2+x^2 \text {, where } \overrightarrow{Q P}=\left(\begin{array}{c}
-1 \\
2 \\
1
\end{array}\right)-\left(\begin{array}{c}
-\frac{7}{2} \\
-\frac{1}{2} \\
-\frac{3}{2}
\end{array}\right)=\left(\begin{array}{c}
\frac{5}{2} \\
\frac{5}{2} \\
\frac{5}{2}
\end{array}\right)=\frac{5}{2}\left(\begin{array}{l}
1 \\
1 \\
1
\end{array}\right)
$

$|\overrightarrow{Q P}|=\left|\frac{5}{2}\left(\begin{array}{l}1 \\ 1 \\ 1\end{array}\right)\right|=\frac{5}{2} \sqrt{1^2+1^2+1^2}=\frac{5}{2} \sqrt{3}$
$x=\frac{\sqrt{3}}{2}$