Let the equation of the plane passing through the line $x-2y-z-5=0=x+y+3z-5$ and parallel to the line $x+y+2z-7=0=2x+3y+z-2$ be $ax+by+cz=65$. Then the distance of the point $(a, b, c)$ from the plane $2x+2y-z+16=0$ is $..........$.

$A$ plane is making intercepts $2, 3, 4$ on $X, Y$ and $Z$-axes respectively. Another plane is passing through the point $(-1, 6, 2)$ and is perpendicular to the line joining the points $(1, 2, 3)$ and $(-2, 3, 4)$. Then the angle between the two planes is

Let a unit vector $\hat{OP}$ make angles $\alpha, \beta, \gamma$ with the positive directions of the coordinate axes $OX, OY, OZ$ respectively,where $\beta \in (0, \frac{\pi}{2})$. If $\hat{OP}$ is perpendicular to the plane passing through the points $(1, 2, 3)$,$(2, 3, 4)$,and $(1, 5, 7)$,then which one of the following is true?

Let the line passing through the points $P(2, -1, 2)$ and $Q(5, 3, 4)$ meet the plane $x - y + z = 4$ at the point $R$. Then the distance of the point $R$ from the plane $x + 2y + 3z + 2 = 0$ measured parallel to the line $\frac{x - 7}{2} = \frac{y + 3}{2} = \frac{z - 2}{1}$ is equal to

Let a line $L$ passing through the point $P(1, 1, 1)$ be perpendicular to the lines $\frac{x-4}{4}=\frac{y-1}{1}=\frac{z-1}{1}$ and $\frac{x-17}{1}=\frac{y-71}{1}=\frac{z}{0}$. Let the line $L$ intersect the $yz$-plane at the point $Q$. Another line parallel to $L$ and passing through the point $S(1, 0, -1)$ intersects the $yz$-plane at the point $R$. Then the square of the area of the parallelogram $PQRS$ is equal to . . . . . . .

The length (in units) of the projection of the line segment,joining the points $(5,-1,4)$ and $(4,-1,3)$,on the plane $x+y+z=7$ is