The coordinates of the foot of the perpendicular from the point $(0,2,3)$ on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$ are

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 shortest distance between the lines $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and $\frac{x + 2}{- 1} = \frac{y - 4}{8} = \frac{z - 5}{4}$ lies in the interval

Point $A$ lies at a distance of $6$ units from the point $(1, 0, 1)$ on the line $\frac{x - 1}{2} = \frac{y}{2} = \frac{z - 1}{1}$ in the negative $z$-direction. Find the coordinates of $A$.

Find the position vector of the image of the point with position vector $\vec{P} = 2\hat{i} + \hat{j} + 3\hat{k}$ in the line whose vector equation is $\vec{r} = \hat{j} - 2\hat{k} + \lambda(\hat{i} + \hat{j} - \hat{k})$.

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.