The angle between the lines $\vec{r}=(2 \hat{i}+\hat{j}-3 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$ and $\frac{x-1}{1}=\frac{y+2}{3}=\frac{z-3}{2}$ is

Let $(\alpha, \beta, \gamma)$ be the coordinates of the foot of the perpendicular drawn from the point $(5, 4, 2)$ on the line $\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k})$. Then the length of the projection of the vector $\alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$ on the vector $6\hat{i} + 2\hat{j} + 3\hat{k}$ is:

Let the point $(-1, \alpha, \beta)$ lie on the line of the shortest distance between the lines $\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$ and $\frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0}$. Then $(\alpha-\beta)^2$ is equal to ....................

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:

The equation of the straight line passing through the points $(4, -5, -2)$ and $(-1, 5, 3)$ is

Let $(\alpha, \beta, \gamma)$ be the foot of the perpendicular from the point $(1, 2, 3)$ on the line $\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}$. Then $19(\alpha + \beta + \gamma)$ is equal to: