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

Consider the line $L$ given by the equation $\frac{x-3}{2}=\frac{y-1}{1}=\frac{z-2}{1}$. Let $Q$ be the mirror image of the point $P_0(2,3,-1)$ with respect to $L$. Let a plane $P$ be such that it passes through $Q$,and the line $L$ is perpendicular to $P$. Then which of the following points is on the plane $P$?

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

$A$ line $L$ passes through the points $\hat{i}+2 \hat{j}+\hat{k}$ and $-2 \hat{i}+3 \hat{k}$. $A$ plane $P$ passes through the origin and the points $4 \hat{k}, 2 \hat{i}+\hat{j}$. The point where the line $L$ meets the plane $P$ is

Let $P$ be a plane containing the line $\frac{x-1}{3}=\frac{y+6}{4}=\frac{z+5}{2}$ and parallel to the line $\frac{x-3}{4}=\frac{y-2}{-3}=\frac{z+5}{7}$. If the point $(1, -1, \alpha)$ lies on the plane $P$,then the value of $|5\alpha|$ is equal to ....... .

From a point $P(\lambda, \lambda, \lambda)$,perpendiculars $PQ$ and $PR$ are drawn respectively on the lines $y=x, z=1$ and $y=-x, z=-1$. If $P$ is such that $\angle QPR$ is a right angle,then the possible value$(s)$ of $\lambda$ is(are)