The line $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{-1}$ intersects the curve $xy = c^2, z = 0$ if $c$ is equal to

The angle between the line $\bar{r}=(\hat{i}+2\hat{j}-\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$ and the plane $\bar{r} \cdot (2\hat{i}-\hat{j}+\hat{k})=4$ is:

The plane containing the line $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and parallel to the line $\frac{x}{1} = \frac{y}{1} = \frac{z}{4}$ passes through the point

The shortest distance between the $z$-axis and the line $x + y + 2z - 3 = 0 = 2x + 3y + 4z - 4$ is

The coordinates of the point where the line $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+3}{4}$ meets the plane $2x+4y-z=1$ are

The equation of the plane passing through the line of intersection of the planes $\overline{r} \cdot(2 \hat{i}-3 \hat{j}+4 \hat{k})=1$ and $\overline{r} \cdot(\hat{i}-\hat{j})+4=0$,and perpendicular to the plane $\overline{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})+8=0$,is given by $\overline{r} \cdot(-5 \hat{i}+2 \hat{j}+12 \hat{k})=\mu$. Then the value of $\mu$ is: