Find the distance of the point $(1, -2, 3)$ from the plane $x - y + z = 5$ measured parallel to the line $\frac{x}{2} = \frac{y}{3} = \frac{z - 1}{-6}$.

Find the equation of the plane passing through the line of intersection of the planes $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=1$ and $\vec{r} \cdot(2 \hat{i}+3 \hat{j}-\hat{k})+4=0$ and parallel to the $x$-axis.

If $\bar{r} \cdot(2 \bar{i}+3 \bar{j}+4 \bar{k})=5$ and $\bar{r} \cdot(\bar{i}+\bar{j}-\bar{k})=7$ are two planes and $(16, -9, 0)$ is a point common to both the planes, then the vector equation of the line of intersection of the planes is $\bar{r}=$

The $XOZ$ plane divides the line segment joining the points $(2, 3, 1)$ and $(6, 7, 1)$ in the ratio: (in $:7$)

The equation of the plane passing through the intersection of the planes $x+y+z=1$ and $2x+3y-z+4=0$ and parallel to the $X$-axis is

$A$ plane $E$ is perpendicular to the two planes $2x - 2y + z = 0$ and $x - y + 2z = 4$,and passes through the point $P(1, -1, 1)$. If the distance of the plane $E$ from the point $Q(a, a, 2)$ is $3\sqrt{2}$,then $(PQ)^2$ is equal to