The values of $p$ and $q$ such that the line joining the points $(7, p, 2)$ and $(q, -2, 5)$ is parallel to the line joining the points $(2, -3, 5)$ and $(-6, -15, 11)$ are:

The foot of the perpendicular from the point $A(2, 0, 5)$ to the line $\frac{x+1}{2} = \frac{y-1}{5} = \frac{z+1}{-1}$ is $P(\alpha, \beta, \gamma)$. Then,which of the following is $NOT$ correct?

Find the angle between the two lines $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$ and $\frac{x-1}{10}=\frac{y+3}{2}=\frac{z+4}{-11}$.

The distance of the point having position vector $-\hat{i} + 2\hat{j} + 6\hat{k}$ from the straight line passing through the point $(2, 3, -4)$ and parallel to the vector $6\hat{i} + 3\hat{j} - 4\hat{k}$ is

The position vector of the point of intersection of the line joining the points $2 \hat{i}-\hat{j}+6 \hat{k}$ and $3 \hat{i}-\hat{j}-7 \hat{k}$, and the line joining the points $2 \hat{i}+\hat{j}-6 \hat{k}$ and $3 \hat{i}-\hat{j}-7 \hat{k}$ is:

Let $a, b \in R$. If the mirror image of the point $P(a, 6, 9)$ with respect to the line $\frac{x-3}{7} = \frac{y-2}{5} = \frac{z-1}{-9}$ is $(20, b, -a-9)$,then $|a+b|$ is equal to