The equation of the line of intersection of the planes $4x + 4y - 5z = 12$ and $8x + 12y - 13z = 32$ can be written as

If the line joining the points $\overline{i} + 2\overline{j}$ and $\overline{j} - 2\overline{k}$ intersects the plane passing through the points $2\overline{i} - \overline{j}$,$2\overline{j} + 3\overline{k}$,and $\overline{k} - 2\overline{i}$ at $\overline{r}$,then $\overline{r} \cdot (\overline{i} + \overline{j} + \overline{k}) = $

Find the vector equation of the plane passing through the intersection of the planes $\vec{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=7$ and $\vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})=9$ and passing through the point $(2,1,3).$

In what ratio does the $xy$-plane divide the line segment joining the points $(1, 2, 3)$ and $(4, 2, 1)$?

Let a line $L$ passing through the point $P(1, 1, 1)$ be perpendicular to the lines $\frac{x-4}{4}=\frac{y-1}{1}=\frac{z-1}{1}$ and $\frac{x-17}{1}=\frac{y-71}{1}=\frac{z}{0}$. Let the line $L$ intersect the $yz$-plane at the point $Q$. Another line parallel to $L$ and passing through the point $S(1, 0, -1)$ intersects the $yz$-plane at the point $R$. Then the square of the area of the parallelogram $PQRS$ is equal to . . . . . . .

The distance of the point $(1, 3, -7)$ from the plane passing through the point $(1, -1, -1)$ and having a normal perpendicular to both the lines $\frac{x - 1}{1} = \frac{y + 2}{-2} = \frac{z - 4}{3}$ and $\frac{x - 2}{2} = \frac{y + 1}{-1} = \frac{z + 7}{-1}$ is . . . .