Question

Sol. [Key Concept: Direction is the cross product of the two vectors; the foot $M$ satisfies $OM \cdot \ d = 0$.]

Direction vector of line is

$$= \left| \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 1 & 2 & 2 \\ 2 & 2 & 1 \end{matrix} \right| = - 2\widehat{i} + 3\widehat{j} - 2\widehat{k}$$

$$\begin{matrix} & \ \\ \frac{x - 1}{2} = & \frac{y - 1}{- 3} = \frac{z - 1}{2} \end{matrix}$$

So equation of line is

$\frac{x - 1}{2} = \frac{y - 1}{- 3} = \frac{z - 1}{2} = \lambda$ let

let ( $2\lambda + 1, - 3\lambda + 1,2\lambda + 1$ ) be the foot of perpendicular from $(0,0,0)$ upon given line

$${\Rightarrow \overrightarrow{OM} \cdot (2\widehat{i} - 3j + 2\widehat{k}) = 0 }{\Rightarrow 2(2\lambda + 1) - 3( - 3\lambda + 1) + 2(2\lambda + 1) = 0 }{\Rightarrow \lambda = - \frac{1}{17}}$$

So $(a,b,c) \equiv \left( \frac{- 2}{17} + 1,\frac{3}{17} + 1,\frac{- 2}{17} + 1 \right)$

$$\Rightarrow 34(a + b + c) = 100 $$