Let a line with direction ratios $\mathrm{a},-4\mathrm{a},-7$ be perpendicular to the lines with direction ratios $3,-1, 2\mathrm{b}$  and$\mathrm{b}, \mathrm{a},-2$. If the point of intersection of the Iine $\frac{\mathrm{x}+1}{{\mathrm{a}}^2+{\mathrm{b}}^2}=\frac{\mathrm{y}-2}{{\mathrm{a}}^2-{\mathrm{b}}^2}=\frac{\mathrm{z}}{1}$ and the plane $\mathrm{x}-\mathrm{y}+\mathrm{z}=0$ is $(\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma })$ then $\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }$ is equal to__________.

$<\mathrm{a},-4\mathrm{a},-7>\perp \mathrm{r}<3,-1,2\mathrm{b}>$
$\begin{array}{l}\Rightarrow 3\mathrm{a}+4\mathrm{a}-14\mathrm{b}=0\Rightarrow 7\mathrm{a}=14\mathrm{b}\Rightarrow \mathrm{a}=2\mathrm{b}\\ <\mathrm{a},-4\mathrm{a},-7>\perp \mathrm{r}<\mathrm{b},\mathrm{a},-2>\\ \Rightarrow \mathrm{ab}-4{\mathrm{a}}^2+14=0\Rightarrow 2{\mathrm{b}}^2-4\times 4{\mathrm{b}}^2+14=0\\ \Rightarrow -14{\mathrm{b}}^2=-14\Rightarrow {\mathrm{b}}^2=1\therefore {\mathrm{a}}^2=4{\mathrm{b}}^2=4\end{array}$
$\therefore$Given line $\frac{\mathrm{x}+1}{{\mathrm{a}}^2+{\mathrm{b}}^2}=\frac{\mathrm{y}-2}{{\mathrm{a}}^2-{\mathrm{b}}^2}=\frac{\mathrm{z}}{1}$ becomes $\frac{\mathrm{x}+1}{5}=\frac{\mathrm{y}-2}{3}=\frac{\mathrm{z}}{1}=\mathrm{k} (\mathrm{say})$
Any point i.e., $(\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma })$on this line can be $(5\mathrm{k}-1,3\mathrm{k}+2,\mathrm{k})$
Also, $(\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma })$ lies on $\mathrm{x}-\mathrm{y}+\mathrm{z}=0$
$\begin{array}{l}\therefore 5\mathrm{k}-1-3\mathrm{k}-2+\mathrm{k}=0\Rightarrow 3\mathrm{k}=3\Rightarrow \mathrm{k}=1\\ \therefore \mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }=9\mathrm{k}+1=9+1=10\end{array}$