The line of intersection of the planes $\vec r \cdot (3\hat i - \hat j + \hat k) = 1$ and $\vec r \cdot (\hat i + 4\hat j - 2\hat k) = 2$ is:

The equation of the plane containing the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and the point $(0,7,-7)$ is

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

Let $L$ be a line passing through the points $2 \hat{i}+3 \hat{j}+8 \hat{k}$ and $\hat{i}+6 \hat{j}+4 \hat{k}$. Let $P$ be a plane passing through $-5 \hat{i}+19 \hat{j}-14 \hat{k}$ and parallel to the vectors $\hat{i}-\hat{j}+\hat{k}$ and $\hat{i}-2 \hat{j}+3 \hat{k}$. If $L$ meets the plane $P$ at a point $A$, then the position vector of $A$ is:

The line of shortest distance between the lines $\frac{x-2}{0}=\frac{y-1}{1}=\frac{z}{1}$ and $\frac{x-3}{2}=\frac{y-5}{2}=\frac{z-1}{1}$ makes an angle of $\cos ^{-1}\left(\sqrt{\frac{2}{27}}\right)$ with the plane $P: ax-y-z=0$,$(a>0)$. If the image of the point $(1,1,-5)$ in the plane $P$ is $(\alpha, \beta, \gamma)$,then $\alpha+\beta-\gamma$ is equal to $........$

The equation of the plane passing through the point $(2,2,1)$ and the intersection of the planes $x+2y-3z+1=0$ and $3x-2y+4z+3=0$ is