Find the equation of the plane passing through the line of intersection of the planes $x+y+z=1$ and $2x+3y+4z=5$ which is perpendicular to the plane $x-y+z=0$.

$A$ plane $E$ is perpendicular to the two planes $2x - 2y + z = 0$ and $x - y + 2z = 4$,and passes through the point $P(1, -1, 1)$. If the distance of the plane $E$ from the point $Q(a, a, 2)$ is $3\sqrt{2}$,then $(PQ)^2$ is equal to

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

The equation of the plane through the intersection of the planes $x+y+z=1$ and $2x+3y-z+4=0$ and parallel to the $x$-axis is:

If $Q(0, -1, -3)$ is the image of the point $P$ in the plane $3x - y + 4z = 2$ and $R$ is the point $(3, -1, -2)$,then the area (in square units) of $\Delta PQR$ is

The equation of the plane passing through the origin and perpendicular to the line $x = 2y = 3z$ is