The equation of the plane passing through the points $(3, 2, 2)$ and $(1, 0, -1)$ and parallel to the line $\frac{x - 1}{2} = \frac{y - 1}{-2} = \frac{z - 2}{3}$ is:

If the three planes $x = 5, 2x - 5ay + 3z - 2 = 0$ and $3bx + y - 3z = 0$ contain a common line,then $(a, b)$ is equal to

Perpendiculars are drawn from points on the line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}$ to the plane $x+y+z=3$. The feet of the perpendiculars lie on the line:

Let the mirror image of the point $(a, b, c)$ with respect to the plane $3x - 4y + 12z + 19 = 0$ be $(a - 6, \beta, \gamma)$. If $a + b + c = 5$,then $7\beta - 9\gamma$ is equal to

If the distance of the point $P(43, \alpha, \beta), \beta < 0$, from the line $\vec{r} = 4\hat{i} - \hat{k} + \mu(2\hat{i} + 3\hat{k}), \mu \in R$ along a line with direction ratios $3, -1, 0$ is $13\sqrt{10}$, then $\alpha^{2} + \beta^{2}$ is equal to . . . . . . .

Let $\gamma \in R$ be such that the lines $L_1: \frac{x+11}{1}=\frac{y+21}{2}=\frac{z+29}{3}$ and $L_2: \frac{x+16}{3}=\frac{y+11}{2}=\frac{z+4}{\gamma}$ intersect. Let $R_1$ be the point of intersection of $L_1$ and $L_2$. Let $O=(0,0,0)$,and $\hat{n}$ denote a unit normal vector to the plane containing both the lines $L_1$ and $L_2$. Match each entry in $List-I$ to the correct entry in $List-II$.

$List-I$	$List-II$
$(P) \gamma$ equals	$(1) -\hat{i}-\hat{j}+\hat{k}$
$(Q)$ $A$ possible choice for $\hat{n}$ is	$(2) \sqrt{\frac{3}{2}}$
$(R) \vec{OR_1}$ equals	$(3) 1$
$(S)$ $A$ possible value of $\vec{OR_1} \cdot \hat{n}$ is	$(4) \frac{1}{\sqrt{6}} \hat{i}-\frac{2}{\sqrt{6}} \hat{j}+\frac{1}{\sqrt{6}} \hat{k}$
	$(5) \sqrt{\frac{2}{3}}$