Equation of the plane which passes through the point of intersection of lines $\frac{x - 1}{3} = \frac{y - 2}{1} = \frac{z - 3}{2}$ and $\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ and has the largest distance from the origin is

The angle between the line with the direction ratios $(2, 5, 1)$ and the plane $8x + 2y - z = 4$ is

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	$(4)$ $\sqrt{2}$
	$(5)$ $\frac{1}{\sqrt{2}}$

Let $P_{1}: \vec{r} \cdot(2 \hat{i} + \hat{j} - 3 \hat{k}) = 4$ be a plane. Let $P_{2}$ be another plane which passes through the points $(2, -3, 2)$,$(2, -2, -3)$,and $(1, -4, 2)$. If the direction ratios of the line of intersection of $P_{1}$ and $P_{2}$ are $16, \alpha, \beta$,then the value of $\alpha + \beta$ is equal to

The coordinates of the point,where the line $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+5}{4}$ meets the plane $2x+4y-z=3$,are

The distance of the point $(-1, 2, -2)$ from the line of intersection of the planes $2x + 3y + 2z = 0$ and $x - 2y + z = 0$ is: