If the plane $7x + 11y + 13z = 3003$ meets the coordinate axes at $A, B, C$,then the centroid of the $\triangle ABC$ is

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 coordinates of the foot of the perpendicular drawn from the origin to the plane $2x - 3y + 4z = 29$ are

If $S$ is the set of all real values of $a$ such that a plane passing through the points $(-a^2, 1, 1), (1, -a^2, 1), (1, 1, -a^2)$ also passes through the point $(-1, -1, 1)$,then $S=$

The equation of the plane in Cartesian form, which is at a distance of $\frac{6}{\sqrt{29}}$ from the origin and its normal vector drawn from the origin being $2 \hat{i}-3 \hat{j}+4 \hat{k}$, is

The equation of the plane in normal form which passes through the points $(-2,1,3), (1,1,1)$ and $(2,3,4)$ is