Find the equation of the plane passing through the point $(-1, 3, 2)$ and perpendicular to each of the planes $x + 2y + 3z = 5$ and $3x + 3y + z = 0$.

Let the plane $ax + by + cz = d$ pass through $(2, 3, -5)$ and be perpendicular to the planes $2x + y - 5z = 10$ and $3x + 5y - 7z = 12$. If $a, b, c, d$ are integers,$d > 0$,and $\text{gcd}(|a|, |b|, |c|, d) = 1$,then the value of $a + 7b + c + 20d$ is equal to

$A$ variable plane is at a distance of $6$ units from the origin. If it meets the coordinate axes in $A, B$,and $C$,then the equation of the locus of the centroid of the $\triangle ABC$ is

The vector equation of a plane,which is at a distance of $8$ units from the origin and which is normal to the vector $\vec{n} = 2\hat{i} + \hat{j} + 2\hat{k}$,is

If a plane passing through the points $(2,3,0), (0,-5,2)$ and $(-2,0,3)$ meets the $X, Y, Z$-axes in $A, B, C$ respectively,then $A=$

$A$ plane meets the coordinate axes at points $A, B$,and $C$ such that the centroid of $\Delta ABC$ is $(1, 2, 3)$. The equation of the plane is