Points on the line $\frac{x+2}{3} = \frac{y+1}{2} = \frac{z-3}{2}$ at a distance of 5 units from the point $(1, 3, 3)$, are

The angle between 2 planes $4x + 8y + z - 8 = 0$ and $y + z - 4 = 0$ is :

Distance of the point $(2, 4, -1)$ from the line $\frac{10+2x}{2} = \frac{y+3}{4} = \frac{6-z}{9}$ is

Let the equation of lines be as $L_1: \vec{r_1} = \vec{a_1} + \lambda\vec{b_1}$ and $L_2: \vec{r_2} = \vec{a_2} + \lambda\vec{b_2}$ such that $\vec{a_1} - \vec{a_2} = 2\hat{i} + 4\hat{j} + 4\hat{k}$ and $\vec{b_1} \times \vec{b_2} = 8\hat{i} - 4\hat{k}$. Then the shortest distance between $L_1$ and $L_2$ is

The vector equation of line passing through (2, -1, 3) and perpendicular to the lines

$\frac{x-2}{3} = \frac{y-1}{1} = \frac{z+2}{2}$ and $\frac{x+3}{-4} = \frac{y-5}{-3} = \frac{z+1}{2}$ is

(Here $\lambda$ is a parameter)

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

(A) $\frac{x-2}{2} = \frac{y+1}{3} = \frac{z-3}{-2}$

(B) $\frac{x-4}{3} = \frac{y-2}{-2} = \frac{z-1}{2}$

(C) $\vec{r} = (2\hat{i}-\hat{j}+3\hat{k}) + \lambda(2\hat{i}+3\hat{j}-2\hat{k})$

(D) $\frac{x-4}{-2} = \frac{y-2}{-3} = \frac{z-1}{2}$

Choose the correct answer from the options given below:

Equation of y-axis in space, in vector form is :