The distance between the parallel lines $\frac{x-1}{2}=\frac{y-2}{-2}=\frac{z-3}{1}$ and $\frac{x}{2}=\frac{y}{-2}=\frac{z}{1}$ is

Let $P(\alpha, \beta, \gamma)$ be the image of the point $Q(3, -3, 1)$ in the line $\frac{x-0}{1} = \frac{y-3}{1} = \frac{z-1}{-1}$ and $R$ be the point $(2, 5, -1)$. If the area of the triangle $PQR$ is $\lambda$ and $\lambda^2 = 14K$,then $K$ is equal to:

Let the line $L$ pass through the point $(-3, 5, 2)$ and make equal angles with the positive coordinate axes. If the distance of $L$ from the point $P(-2, r, 1)$ is $\sqrt{\frac{14}{3}}$,then the sum of all possible values of $r$ is:

The acute angle between the line joining the points $(2, 1, -3)$ and $(-3, 1, 7)$ and a line parallel to $\frac{x - 1}{3} = \frac{y}{4} = \frac{z + 3}{5}$ passing through the point $(-1, 0, 4)$ is

The angle between the lines $\vec{r}=(2 \hat{i}-3 \hat{j}+\hat{k})+\lambda(\hat{i}+4 \hat{j}+3 \hat{k})$ and $\vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(\hat{i}+2 \hat{j}-3 \hat{k})$ is

When are the two lines $x = ay + b, z = cy + d$ and $x = a'y + b', z = c'y + d'$ perpendicular to each other?