The Cartesian equations of the line passing through the point $A(0, 1, 1)$ and parallel to the $X$-axis are:

Let $(\alpha, \beta, \gamma)$ be the foot of the perpendicular from the point $(1, 2, 3)$ on the line $\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}$. Then $19(\alpha + \beta + \gamma)$ is equal to:

Two lines $L_1: x=5, \frac{y}{3-\alpha}=\frac{z}{-2}$ and $L_2: x=\alpha, \frac{y}{-1}=\frac{z}{2-\alpha}$ are coplanar. Then $\alpha$ can take value$(s)$.

The foot of the perpendicular from the point $A(2, 0, 5)$ to the line $\frac{x+1}{2} = \frac{y-1}{5} = \frac{z+1}{-1}$ is $P(\alpha, \beta, \gamma)$. Then,which of the following is $NOT$ correct?

The point of intersection of the lines $\frac{x - 5}{3} = \frac{y - 7}{-1} = \frac{z + 2}{1}$ and $\frac{x + 3}{-36} = \frac{y - 3}{2} = \frac{z - 6}{4}$ is

The distance of the point $P(-2, 4, -5)$ from the line $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}$ is