$A(1, -2, 1)$ and $B(2, -1, 2)$ are the end points of a line segment. If $D(\alpha, \beta, \gamma)$ is the foot of the perpendicular drawn from $C(1, 2, 3)$ to $AB$,then $\alpha^2 + \beta^2 + \gamma^2 =$

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.

Find the angle between the lines $\vec{r}=3 \hat{i}-2 \hat{j}+6 \hat{k}+\lambda(2 \hat{i}+\hat{j}+2 \hat{k})$ and $\vec{r}=(2 \hat{j}-5 \hat{k})+\mu(6 \hat{i}+3 \hat{j}+2 \hat{k})$.

Let a line $L$ passing through the point $(1, 1, 1)$ be perpendicular to both the vectors $2\hat{i} + 2\hat{j} + \hat{k}$ and $\hat{i} + 2\hat{j} + 2\hat{k}$. If $P(a, b, c)$ is the foot of the perpendicular from the origin on the line $L$, then the value of $34(a + b + c)$ is:

The shortest distance between the lines $1+x=2y=-12z$ and $x=y+2=6z-6$ is

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $x-3=\frac{y-k}{2}=z$ intersect,then the value of $k$ is