The point of intersection of the line $\frac{x - 1}{3} = \frac{y + 2}{4} = \frac{z - 3}{-2}$ and the plane $2x - y + 3z - 1 = 0$ is:

The plane containing the line $\frac{x - 3}{2} = \frac{y + 2}{-1} = \frac{z - 1}{3}$ and also containing its projection on the plane $2x + 3y - z = 5$ contains which one of the following points?

The equation of a plane at a distance of $\sqrt{\frac{2}{21}}$ from the origin,which contains the line of intersection of the planes $x-y-z-1=0$ and $2x+y-3z+4=0$,is:

$A$ line with 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 $Q$. Find the length of the line segment $PQ$.

If the mirror image of the point $(1, 3, 5)$ with respect to the plane $4x - 5y + 2z = 8$ is $(\alpha, \beta, \gamma)$,then $5(\alpha + \beta + \gamma)$ equals:

The square of the distance of the point of intersection of the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{6}$ and the plane $2x-y+z=6$ from the point $(-1,-1,2)$ is .... .