If the image of the point $P(1, -2, 1)$ with respect to the line passing through the points $B(1, 1, 2)$ and $C(2, 2, 1)$ is $R(l, m, n)$, then $l^2 + m^2 + n^2 =$

The Cartesian equation of the line which passes through the points $(3,1,2)$ and $(-1,2,1)$ is

Let the line $L$ intersect the lines $x-2=-y=z-1$ and $2(x+1)=2(y-1)=z+1$,and be parallel to the line $\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}$. Then which of the following points lies on $L$?

The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$ and $l^2+m^2-n^2=0$ is

The point of intersection of the lines represented by $\overline{r}=(\overline{i}-6 \overline{j}+2 \overline{k})+t(\overline{i}+2 \overline{j}+\overline{k})$ and $\overline{r}=(4 \overline{j}+\overline{k})+s(2 \overline{i}+\overline{j}+2 \overline{k})$ is

Let the values of $p$,for which the shortest distance between the lines $\frac{x+1}{3}=\frac{y}{4}=\frac{z}{5}$ and $\overrightarrow{r}=(p\hat{i}+2\hat{j}+\hat{k})+\lambda(2\hat{i}+3\hat{j}+4\hat{k})$ is $\frac{1}{\sqrt{6}}$,be $a$ and $b$ $(a < b)$. Then the length of the latus rectum of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is: