Let the image of the point $P(1, 2, 3)$ in the plane $2x - y + z = 9$ be $Q$. If the coordinates of the point $R$ are $(6, 10, 7)$,then the square of the area of the triangle $PQR$ is $.....$.

Find the equation of the plane passing through the intersection of the planes $x + 2y + 3z = 2$ and $x - y + z = 3$ and at a distance of $\frac{2}{\sqrt{3}}$ from the point $(3, 1, -1)$.

The equation of the plane passing through the point $(2, 5, -3)$ and perpendicular to the planes $x + 2y + 2z = 1$ and $x - 2y + 3z = 4$ is:

Let $P(2,1,5)$ be a point in space and $Q$ be a point on the line $\vec{r}=(\hat{i}-\hat{j}+2\hat{k})+\mu(-3\hat{i}+\hat{j}+5\hat{k})$. Then the value of $\mu$ for which the vector $\vec{PQ}$ is parallel to the plane $3x-y+4z=1$ is

The direction ratios of the normal to the plane passing through the points $(1, 2, -3)$, $(-1, -2, 1)$ and parallel to $\frac{x-2}{2}=\frac{y+1}{3}=\frac{z}{4}$ are:

The foot of the perpendicular drawn from $A(1, 2, 2)$ onto the plane $x+2y+2z-5=0$ is $B(\alpha, \beta, \gamma)$. If $\pi(x, y, z) \equiv x+2y+2z+5=0$ is a plane, then $-\pi(A) : \pi(B) =$ ?