The equation of a plane,containing the line of intersection of the planes $2x - y - 4 = 0$ and $y + 2z - 4 = 0$ and passing through the point $(2, 1, 0)$,is

The equation of the plane passing through $(1, -1, 2)$ and perpendicular to the planes $x + 2y - 2z = 4$ and $3x + 2y + z = 6$ is:

If the cartesian equation of the plane passing through the point $\hat{i}+2 \hat{j}+\hat{k}$ and parallel to the vectors $2 \hat{i}+3 \hat{j}+\hat{k}$ and $-\hat{i}+2 \hat{j}-3 \hat{k}$ is $a x+b y+c z=1$, then $18(a+b+c)$ is equal to

Let the plane passing through the point $(-1, 0, -2)$ and perpendicular to each of the planes $2x + y - z = 2$ and $x - y - z = 3$ be $ax + by + cz + 8 = 0$. Then the value of $a + b + c$ is equal to:

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

If the foot of the perpendicular drawn from the origin to a plane is $P(2,-1,4)$,then the equation of the plane is