The sum of all values of $ \alpha $, for which the shortest distance between the lines $ \frac{x+1}{\alpha}=\frac{y-2}{-1}=\frac{z-4}{-\alpha} $ and $ \frac{x}{\alpha}=\frac{y-1}{2}=\frac{z-1}{2\alpha} $ is $ \sqrt{2} $, is

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

If the direction ratios of a line are in the ratio $2 : 1 : 2$ and it intersects the lines $x = y + a = z$ and $x + a = 2y = 2z$,then the coordinates of the points of intersection are:

The distance of the point $({x_1}, {y_1}, {z_1})$ from the line $\frac{{x - {x_2}}}{l} = \frac{{y - {y_2}}}{m} = \frac{{z - {z_2}}}{n}$,where $l, m, n$ are the direction cosines of the line,is given by:

Let the image of the point $P(0, -5, 0)$ in the line $\frac{x-1}{2} = \frac{y}{1} = \frac{z+1}{-2}$ be the point $R$ and the image of the point $Q(0, -1/2, 0)$ in the line $\frac{x-1}{-1} = \frac{y+9}{4} = \frac{z+1}{1}$ be the point $S$. Then the square of the area of the parallelogram $PQRS$ is . . . . . . .

If $P$ is a point on the line parallel to the vector $2 \hat{i}-3 \hat{j}-6 \hat{k}$ and passing through the point $A$ whose position vector is $\hat{i}+2 \hat{j}-2 \hat{k}$ and $AP=21$, then the position vector of $P$ can be