Find the values of $p$ so that the lines $\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2}$ and $\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles.

The Cartesian equation of a line is $\frac{x+3}{2}=\frac{y-5}{4}=\frac{z+6}{2}$. Find the vector equation for the line.

The angle between the lines $2x = 3y = -z$ and $6x = -y = -4z$ is (in $^{\circ}$)

Let $P(\alpha, \beta, \gamma)$ be the image of the point $Q(3, -3, 1)$ in the line $\frac{x-0}{1} = \frac{y-3}{1} = \frac{z-1}{-1}$ and $R$ be the point $(2, 5, -1)$. If the area of the triangle $PQR$ is $\lambda$ and $\lambda^2 = 14K$,then $K$ is equal to:

The shortest distance between the skew lines $\vec{r}=(-\hat{i}-2 \hat{j}-3 \hat{k})+t(3 \hat{i}-2 \hat{j}-2 \hat{k})$ and $\vec{r}=(7 \hat{i}+4 \hat{k})+s(\hat{i}-2 \hat{j}+2 \hat{k})$ is

If the lines $\frac{1-x}{3}=\frac{7y-14}{2\lambda}=\frac{z-3}{2}$ and $\frac{7-7x}{3\lambda}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles,then $\lambda=$