If the perpendicular distance from $(1, 2, 4)$ to the plane $2x + 2y - z + k = 0$ is $3$,then $k =$

The point which lies on the plane passing through the points $A(\hat{i}-2\hat{j}-3\hat{k})$, $B(3\hat{i}-\hat{j}+4\hat{k})$, and $C(-3\hat{i}+2\hat{j}-5\hat{k})$ is:

$A$ plane ( $\pi$ ) passing through the point $(1, 2, -3)$ is perpendicular to the planes $x + y - z + 4 = 0$ and $2x - y + z + 1 = 0$. If the equation of the plane ( $\pi$ ) is $ax + by + cz + 1 = 0$, then $a^2 + b^2 + c^2 =$

In each of the following cases,determine the direction cosines of the normal to the plane and the distance from the origin: $5y + 8 = 0$.

If $M$ denotes the midpoint of the line segment joining $A(4, 5, -10)$ and $B(-1, 2, 1)$, then the equation of the plane passing through $M$ and perpendicular to $AB$ is:

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: