The vertices B and C of a triangle ABC lie on the line $\frac{x}{1}=\frac{1-y}{-2}=\frac{\mathrm{z}-2}{3}$. The coordinates of A and $B$ are $(1,6,3)$ and $(4,9, \alpha)$ respectively and $C$ is at a distance of 10 units from $B$. The area (in sq. units) of $\triangle A B C$ is :

Let L be the line $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}$ and let S be the set of all points $(\mathrm{a}, \mathrm{b}, \mathrm{c})$ on L , whose distance from the line $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z-9}{0}$ along the line $L$ is 7 . Then $\sum\limits_{(a, b, c) \in S}(a+b+c)$ is equal to :

Let $\mathrm{P}(\alpha, \beta, \gamma)$ be the point on the line $\frac{x-1}{2}=\frac{y+1}{-3}=z$ at a distance $4 \sqrt{14}$ from the point $(1,-1,0)$ and nearer to the origin. Then the shortest distance, between the lines $\frac{x-\alpha}{1}=\frac{y-\beta}{2}=\frac{z-\gamma}{3}$ and $\frac{x+5}{2}=\frac{y-10}{1}=\frac{z-3}{1}$, is equal to

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