The equation of the plane passing through the intersection of the planes $2x - 5y + z = 3$ and $x + y + 4z = 5$ and parallel to the plane $x + 3y + 6z = 1$ is $x + 3y + 6z = k$,where $k$ is:

The image of the point $(-1, 3, 4)$ in the plane $x - 2y = 0$ is

If a line $L$ is common to the planes $x-y+z+2=0$ and $2x+y-2z+5=0$, then the direction cosines of the line $L$ are

$A$ plane $ax+by+cz+1=0$ is perpendicular to the two planes $2x-2y+z=0$ and $x-y+2z=4$ and passes through the point $(1, -2, 1)$. Then $a+b-c=$

The distance of the point $(1,-2,3)$ from the plane $x-y+z=5$ measured parallel to a line,whose direction ratios are $2,3,-6$ is :

Let the line $L: \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{1}$ intersect the plane $2x+y+3z=16$ at the point $P$. Let the point $Q$ be the foot of the perpendicular from the point $R(1, -1, -3)$ on the line $L$. If $\alpha$ is the area of triangle $PQR$,then $\alpha^2$ is equal to $...........$.