The distance between the parallel planes $ax + by + cz + d = 0$ and $ax + by + cz + d' = 0$ is:

If the plane $2x + 3y + 5z = 1$ intersects the coordinate axes at the points $A, B, C$,then the centroid of $\triangle ABC$ is

If $\alpha$ and $\beta$ are scalars and $\vec{r} = (2+\alpha-3\beta) \hat{i} + (\beta-3) \hat{j} + (2\alpha-5\beta-1) \hat{k}$ is the equation of a plane, then its equation in Cartesian form is:

Let $S$ be the set of all real values of $\lambda$ such that a plane passing through the points $(-\lambda^2, 1, 1)$,$(1, -\lambda^2, 1)$,and $(1, 1, -\lambda^2)$ also passes through the point $(-1, -1, 1)$. Then $S$ is equal to

$A$ vector $\overrightarrow{V}$ in the first octant is inclined to the $x$-axis at $60^{\circ}$,to the $y$-axis at $45^{\circ}$ and to the $z$-axis at an acute angle. If a plane passing through the points $(\sqrt{2}, -1, 1)$ and $(a, b, c)$ is normal to $\overrightarrow{V}$,then:

The equation of the plane passing through the point $(1,1,1)$ and perpendicular to the planes $2x-y-2z=5$ and $3x-6y+2z=7$ is