If the foot of the perpendicular drawn from (1, 9, 7) to the line passing through the point (3, 2, 1)
and parallel to the planes x + 2y + z = 0 and $3y-z=3$ is $(\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma })$ , then $\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }$ is equal to

$\begin{array}{l}\mathrm{L}=\frac{\mathrm{x}-3}{\mathrm{\ell }}=\frac{\mathrm{y}-2}{\mathrm{m}}=\frac{\mathrm{z}-1}{\mathrm{n}}\\ {\mathrm{\pi }}_1=\mathrm{x}+2\mathrm{y}+\mathrm{z}=0,{\mathrm{\pi }}_2=3\mathrm{y}-\mathrm{z}=3\\ \mathrm{L}//{\mathrm{\pi }}_1\Rightarrow \mathrm{\ell }+2\mathrm{m}+\mathrm{n}=0\\ \mathrm{L}//{\mathrm{\pi }}_2\Rightarrow 3\mathrm{m}-\mathrm{n}=0\end{array}$

$\begin{array}{l}2\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} 2\\ 3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-1 \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} 3\end{array}$
$\begin{array}{l}\frac{\mathrm{\ell }}{5}=\frac{\mathrm{m}}{-1}=\frac{\mathrm{n}}{-3}\\ \frac{\mathrm{x}-3}{5}=\frac{\mathrm{y}-2}{-1}=\frac{\mathrm{z}-1}{-3}=\mathrm{t}\\ \mathrm{Q}=(5\mathrm{t}+3,2-\mathrm{t},1-3\mathrm{t})\\ \text{ d.r's of }\mathrm{PQ}=(5\mathrm{t}+2,-7-\mathrm{t},-6-3\mathrm{t})\end{array}$
PQ perpendicular $\Rightarrow 5(5t+2)-1(-7-t)-3(-6-3t)$
$\begin{array}{l}=25\mathrm{t}+10+7+\mathrm{t}+18+9\mathrm{t}=0\\ 35\mathrm{t}=-35\Rightarrow \mathrm{t}=-1\\ \mathrm{Q}=(-2,3,4)=(\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma })\\ \mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }=-2+3+4=7-2=5\end{array}$