Let the plane $P : 8x + \alpha_1 y + \alpha_2 z + 12 = 0$ be parallel to the line $L : \frac{x + 2}{2} = \frac{y - 3}{3} = \frac{z + 4}{5}$. If the intercept of $P$ on the $y$-axis is $1$,then the distance between $P$ and $L$ is:

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 equation of the line given by the intersection of planes $x + y + z - 1 = 0$ and $4x + y - 2z + 2 = 0$ in the symmetrical form is represented by which of the following equations?

If the line $\frac{x - 1}{2} = \frac{y + \alpha}{\alpha} = \frac{z - \beta}{2}$ lies in the plane $2x + y + z = 5$,then $\alpha + \beta$ is

The acute angle between the line $\frac{x-5}{2}=\frac{y+1}{-1}=\frac{z+4}{1}$ and the plane $3x-4y-z+5=0$ is

Find the distance of the point $(-1, -2, -1)$ from the plane passing through the point $(1, 1, 1)$ and perpendicular to both lines $L_1: \frac{x-1}{1} = \frac{y-1}{0} = \frac{z-1}{-1}$ and $L_2: \frac{x-1}{0} = \frac{y-1}{1} = \frac{z-1}{-1}$.