The distance of the point $(-1, -5, -10)$ from the point of intersection of the line $\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12}$ and the plane $x-y+z=5$ is

The position vectors of points $A$ and $B$ are $i - j + 3k$ and $3i + 3j + 3k$ respectively. The equation of a plane is $r \cdot (5i + 2j - 7k) + 9 = 0$. The points $A$ and $B$:

Let $\ell_1$ and $\ell_2$ be the lines $\vec{r}_1=\lambda(\hat{i}+\hat{j}+\hat{k})$ and $\vec{r}_2=(\hat{j}-\hat{k})+\mu(\hat{i}+\hat{k})$,respectively. Let $X$ be the set of all the planes $H$ that contain the line $\ell_1$. For a plane $H$,let $d(H)$ denote the smallest possible distance between the points of $\ell_2$ and $H$. Let $H_0$ be the plane in $X$ for which $d(H_0)$ is the maximum value of $d(H)$ as $H$ varies over all planes in $X$. Match each entry in List-$I$ to the correct entries in List-$II$.

List-$I$	List-$II$
$(P)$ The value of $d(H_0)$ is	$(1)$ $\sqrt{3}$
$(Q)$ The distance of the point $(0,1,2)$ from $H_0$ is	$(2)$ $\frac{1}{\sqrt{3}}$
$(R)$ The distance of origin from $H_0$ is	$(3)$ $0$
$(S)$ The distance of origin from the point of intersection of planes $y=z, x=1$ and $H_0$ is	$(4)$ $\sqrt{2}$
	$(5)$ $\frac{1}{\sqrt{2}}$

If $P=(0,1,0)$ and $Q=(0,0,1)$,then the length of the projection of the line segment $PQ$ on the plane $x+y+z=3$ is:

If the lines $L_1 : \frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2}$ and $L_2 : \frac{x - 1}{3k} = \frac{y - 5}{1} = \frac{z - 6}{-5}$ are perpendicular to each other, then the equation of a plane containing the line $L_1$ and parallel to the line $L_2$ for the value of $k$ that satisfies this condition is

The distance between the line $\vec{r} = (2\hat{i} - 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} + 4\hat{k})$ and the plane $\vec{r} \cdot (\hat{i} + 5\hat{j} + \hat{k}) = 5$ is: