The equation of the plane passing through the line $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{4}$ and perpendicular to the plane $x+2y+z=12$ is given by $ax+by+cz+4=0$. Then:

In what ratio does the $xy$-plane divide the line segment joining the points $(1, 2, 3)$ and $(4, 2, 1)$?

$A$ line $L$ passes through the points $(1, 2, -3)$ and $(3, 3, -1)$ and a plane $\pi$ passes through the points $(2, 1, -2), (-2, -3, 6)$,and $(0, 2, -1)$. If $\theta$ is the angle between the line $L$ and plane $\pi$,then $27 \cos^2 \theta = $

The distance of the point $(1, -2, 3)$ from the plane $x - y + z = 5$ measured parallel to the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{-6}$ is

$A$ line with direction ratios $1, -1, 2$ intersects the lines $\frac{x}{2} = \frac{y}{3} = \frac{z+1}{3}$ and $\frac{x+1}{-1} = \frac{y-2}{1} = \frac{z}{4}$ at the points $P$ and $Q$, respectively. If the length of the line segment $PQ$ is $\alpha$, then $225\alpha^2$ is equal to:

Three lines $L_1: \overrightarrow{r} = \lambda \hat{i}, \lambda \in R$,$L_2: \overrightarrow{r} = \hat{k} + \mu \hat{j}, \mu \in R$,and $L_3: \overrightarrow{r} = \hat{i} + \hat{j} + v\hat{k}, v \in R$ are given. For which point$(s)$ $Q$ on $L_2$ can we find a point $P$ on $L_1$ and a point $R$ on $L_3$ such that $P, Q,$ and $R$ are collinear?