If $A(1, 0, 2)$,$B(2, 1, 0)$,$C(2, -5, 3)$,and $D(0, 3, 2)$ are four points and the point of intersection of the lines $AB$ and $CD$ is $P(a, b, c)$,then $a + b + c =$

Statement $-1$: The point $A(1, 0, 7)$ is the mirror image of the point $B(1, 6, 3)$ in the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$.
Statement $-2$: The line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ bisects the line segment joining $A(1, 0, 7)$ and $B(1, 6, 3)$.

Find the coordinates of the point where the line passing through $(5, 1, 6)$ and $(3, 4, 1)$ crosses the $YZ$-plane.

Let $P$ be the image of the point $Q(7,-2,5)$ in the line $L: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ and $R(5, p, q)$ be a point on $L$. Then the square of the area of $\triangle P Q R$ is $\qquad$

If the image of the point $P(1, 2, a)$ in the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{7-z}{2}$ is $Q(5, b, c)$, then $a^{2}+b^{2}+c^{2}$ is equal to

The straight line $\frac{x-3}{3}=\frac{y-2}{1}=\frac{z-1}{0}$ is