Let ℓ 1 and ℓ 2 be the lines r → 1 = λ i ^ + j ^ + k ^ and r → 2 = j ^ - k ^ +…

Given,

$H_0$ will be the plane containing the line ${\ell }_1$ and parallel to ${\ell }_2$.

So, the normal vector of plane parallel to ${\ell }_1$ and ${\ell }_2$ is given by,

$\left|\begin{array}{lll}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 1& 0& 1\end{array}\right|=\hat{j}\left(1\right)-\hat{j}\left(1-1\right)+\hat{k}\left(-1\right)$$=\hat{i}-\hat{k}$

Hence, the equation of plane $H_{0 }$ will be,

$H_{0 }: x-z=C$ which passes through origin,

So,$C=0$

$\therefore H_0 : x-z=0$

Now solving,

$\left(\mathrm{P}\right)$ $d\left(H_0\right)=1$ distance of point $\left(0, 1, -1\right)$ from $H$.

$d=\left|\frac{0-\left(-1\right)}{\sqrt{2}}\right|=\frac{1}{\sqrt{2}} \therefore P\to 5$

$\left(\mathrm{Q}\right)$ $d=\left|\frac{0-2}{\sqrt{2}}\right|=\sqrt{2} \therefore Q\to 4$

$\left(\mathrm{R}\right)$ $d=\left|\frac{0}{\sqrt{2}}\right|=0 \therefore R\to 3$

$\left(\mathrm{S}\right)$ Point of intersection will be of given planes$y=z, x=1 \& x-z=0$  will be, $\left(1, 1, 1\right)$

Hence, distance $d=\sqrt{1+1+1}=\sqrt{3}$ $\therefore S\to 1$

$\therefore$ Option (B) is correct.