The direction ratios of the line of intersection of the planes $x-y+z-5=0$ and $x-3y-6=0$ are:

The equation of the plane passing through the points $(3, 2, 2)$ and $(1, 0, -1)$ and parallel to the line $\frac{x - 1}{2} = \frac{y - 1}{-2} = \frac{z - 2}{3}$ is:

On which of the following lines does the point of intersection of the line $\frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}$ and the plane $x+y+z=2$ lie?

If the foot of the perpendicular from the point $A(-1, 4, 3)$ on the plane $P: 2x + my + nz = 4$ is $B\left(-2, \frac{7}{2}, \frac{3}{2}\right)$,then the distance of the point $A$ from the plane $P$,measured parallel to a line with direction ratios $3, -1, -4$,is equal to.

The equation of the plane passing through the intersection of the planes $x + 2y + z - 1 = 0$ and $2x + y + 3z - 2 = 0$ is perpendicular to the plane $x + y + z - 1 = 0$. If this plane is also parallel to the plane $x + ky + 3z - 1 = 0$,then the value of $k$ is:

Let the plane $2x + 3y + z + 20 = 0$ be rotated through a right angle about its line of intersection with the plane $x - 3y + 5z = 8$. If the mirror image of the point $(2, -1/2, 2)$ in the rotated plane is $B(a, b, c)$,then: