$X$ intercept of the plane containing the line of intersection of the planes $x-2y+z+2=0$ and $3x-y-z+1=0$ and also passing through $(1,1,1)$ is

The distance of the point $2i + j - k$ from the plane $r \cdot (i - 2j + 4k) = 9$ 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.

If $P=(0,1,0)$ and $Q=(0,0,1)$,then the length of the projection of the line segment $PQ$ on the plane $x+y+z=3$ is:

Let $P_1$ be the plane $3x - y - 7z = 11$ and $P_2$ be the plane passing through the points $(2, -1, 0)$,$(2, 0, -1)$,and $(5, 1, 1)$. If the foot of the perpendicular drawn from the point $(7, 4, -1)$ on the line of intersection of the planes $P_1$ and $P_2$ is $(\alpha, \beta, \gamma)$,then $\alpha + \beta + \gamma$ is equal to $............$.

The distance of the point $(3, 4, 5)$ from the point of intersection of the line $\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2}$ and the plane $x+y+z=2$ is: (in $units$)