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 distance between the line $\bar{r} = 3\hat{i} - 2\hat{j} + \hat{k} + \lambda(\hat{i} - 2\hat{j})$ and the plane $\bar{r} \cdot (2\hat{i} + \hat{j} + \hat{k}) = 4$ is

If the angle $\theta$ between the line $\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$ and the plane $2x-y+\sqrt{\lambda}z+4=0$ is such that $\sin \theta=\frac{1}{3}$,then $\lambda+1=$

The equation of a plane containing the line of intersection of the planes $2x - y - 4 = 0$ and $y + 2z - 4 = 0$ and passing through the point $(1, 1, 0)$ is

$A$ plane which passes through the point $(3, 2, 0)$ and the line $\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}$ is

The distance of the point $(-1, -5, -10)$ from the point of intersection of the line $\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12}$ and the plane $x-y+z=5$ is