If the line $\frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n}$ is parallel to the plane $ax + by + cz + d = 0$,then:

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

Let $L_1$ and $L_2$ be the following straight lines:
$L_1: \frac{x-1}{1} = \frac{y}{-1} = \frac{z-1}{3}$ and $L_2: \frac{x-1}{-3} = \frac{y}{-1} = \frac{z-1}{1}$.
Suppose the straight line $L: \frac{x-\alpha}{l} = \frac{y-1}{m} = \frac{z-\gamma}{-2}$ lies in the plane containing $L_1$ and $L_2$,and passes through the point of intersection of $L_1$ and $L_2$. If the line $L$ bisects the acute angle between the lines $L_1$ and $L_2$,then which of the following statements is/are $TRUE$?
$(A)$ $\alpha-\gamma=3$
$(B)$ $l+m=2$
$(C)$ $\alpha-\gamma=1$
$(D)$ $l+m=0$

The distance of the point having position vector $\hat{i}-2 \hat{j}-6 \hat{k}$ from the straight line passing through the point $(2, -3, -4)$ and parallel to the vector $6 \hat{i}+3 \hat{j}-4 \hat{k}$ is units.

The distance of the point of intersection of the line $\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z - 5}{2}$ and the plane $x + y + z = 17$ from the point $(3, 4, 5)$ is given by

The line of intersection of the planes $r \cdot (i - 3j + k) = 1$ and $r \cdot (2i + 5j - 3k) = 2$ is parallel to which vector?