What is the point of intersection of the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z + 2}{3}$ and the plane $2x + 3y + z = 0$?

The distance of the point $A(3, -4, 5)$ from the plane $2x + 5y - 6z = 16$ measured along the line $\frac{x}{2} = \frac{y}{1} = \frac{z}{-2}$ is

The line $\ell$ passes through the point $P(2, 1, 1)$ and is parallel to the plane $x + y + 2z = 18$. If line $\ell$ intersects the line $L_2: \frac{x + 2}{3} = \frac{y + 1}{-1} = \frac{z - 2}{1}$, then the equation of the line $\ell$ is:

$A$ line $l$ passes through the origin and is perpendicular to the lines $l_1 = (3 + t)\hat{i} + (-1 + 2t)\hat{j} + (4 + 2t)\hat{k}$ and $l_2 = (3 + 2s)\hat{i} + (3 + 2s)\hat{j} + (2 + s)\hat{k}$.
Statement $1$: Line $l$ and $l_2$ are coplanar lines.
Statement $2$: Line $l$ and $l_2$ are intersecting lines.

Let $P (-2,-1,1)$ and $Q \left(\frac{56}{17}, \frac{43}{17}, \frac{111}{17}\right)$ be the vertices of the rhombus $PRQS$. If the direction ratios of the diagonal $RS$ are $\alpha, -1, \beta$,where both $\alpha$ and $\beta$ are integers of minimum absolute values,then $\alpha^{2}+\beta^{2}$ is equal to $.....$

Let a line $L_1$ pass through the origin and be perpendicular to the lines $L_2: \vec{r} = (3+t)\hat{i} + (2t-1)\hat{j} + (2t+4)\hat{k}$ and $L_3: \vec{r} = (3+2s)\hat{i} + (3+2s)\hat{j} + (2+s)\hat{k}$, where $t, s \in R$. If $(a, b, c)$, with $a \in Z$, is the point on $L_3$ at a distance of $\sqrt{17}$ from the point of intersection of $L_1$ and $L_2$, then $(a+b+c)^2$ is equal to . . . . . . .