If the foot of the perpendicular drawn from (1, 9, 7) to the line...

We know that,

Direction ratio of line parallel the planes $x + 2y + z = 0$ and $3y - z = 3$ is given by,

D . R = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 1 & 2 & 1 \\ 0 & 3 & - 1\end{cases}\right| = - 5\overset{^}{i} + \overset{^}{j} + 3\overset{^}{k}

Now given line is passing through \left(3, 2, 1\right)$and having direction ratio \left(- 5, 1, 3\right)$,

So, equation of line will be $\dfrac{x - 3}{- 5} = \dfrac{y - 2}{1} = \dfrac{z - 1}{3} = t$

Now let any point line is given by $Q ≡ \left(- 5t + 3, t + 2, 3t + 1\right)$

Now we need to find foot of perpendicular from point $P\left(1, 9, 7\right)$,

So direction ratio of line $QP$ will be \left(- 5t + 3 - 1, t + 2 - 9, 3t + 1 - 7\right) ≡ \left(- 5t + 2, t - 7, 3t - 6\right)$</p><p>Now we know that$QP$will be perpendicular to the given line, so by condition of perpendicular lines we get,</p><p>$\left[\left(- 5t + 2\right)\overset{^}{i} + \left(t - 7\right)\overset{^}{j} + \left(3t - 6\right)\overset{^}{k}\right] \cdot \left[- 5\overset{^}{i} + \overset{^}{j} + 3\overset{^}{k}\right] = 0$</p><p>$⇒ 25t - 10 + t - 7 + 9t - 18 = 0$</p><p>$⇒ t = 1$</p><p>So, the point$Q = \left(- 2, 3, 4\right)$, now comparing with \left(\alpha, \beta, \gamma\right)$, we get $\alpha + \beta + \gamma = - 2 + 3 + 4 = 5$