The plane $2x - 3y + 6z - 11 = 0$ makes an angle $\sin^{-1}(\alpha)$ with the $X$-axis. The value of $\alpha$ is equal to:

$A$ line with positive direction cosines passes through the point $P(2, -1, 2)$ and makes equal angles with the coordinate axes. If the line meets the plane $2x + y + z = 9$ at point $Q,$ then the length $PQ$ equals

The plane $lx + my = 0$ is rotated by an angle $\alpha$ about its line of intersection with the plane $z = 0$. Find the equation of the plane in its new position.

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

The angle made by the vector $2\hat{i}-\hat{j}+\hat{k}$ with the plane represented by $\vec{r} \cdot(\hat{i}+\hat{j}+2\hat{k})=7$ is (in $^{\circ}$)

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: