Let $Q$ be the mirror image of the point $P(1, 0, 1)$ with respect to the plane $S: x + y + z = 5$. If a line $L$ passing through $(1, -1, -1)$,parallel to the line $PQ$,meets the plane $S$ at $R$,then $QR^{2}$ is equal to

The plane $4x + 7y + 4z + 81 = 0$ is rotated through a right angle about its line of intersection with the plane $5x + 3y + 10z = 25$. The equation of the plane in its new position is $x - 4y + 6z = k$,where $k$ is:

Let $\bar{A}$ be a vector parallel to the line of intersection of planes $P_1$ and $P_2$ passing through the origin. $P_1$ is parallel to the vectors $2 \hat{j}+3 \hat{k}$ and $4 \hat{j}-3 \hat{k}$,and $P_2$ is parallel to $\hat{j}-\hat{k}$ and $3 \hat{i}+3 \hat{j}$. Then the angle between $\bar{A}$ and $2 \hat{i}+\hat{j}-2 \hat{k}$ is

The angle between the line $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z + 3}{-2}$ and the plane $x + y + 4 = 0$ is ......... $^o$.

If the lines $\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar,then the equation of the plane containing these lines is:

Find the equation of the plane which contains the line of intersection of the planes $\vec{r} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})-4=0$ and $\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})+5=0$ and which is perpendicular to the plane $\vec{r} \cdot(5 \hat{i}+3 \hat{j}-6 \hat{k})+8=0$.