Let P be the plane passing through the line $\frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{-3}=\frac{\mathrm{z}+5}{7}$ and the point (2, 4, -3). if the image of the point 
(-1, 3, 4) in the plane P is $(\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma })\text{ then }\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }$ is equal to

The equation of plane passing through the points (1, 2, -5) and (2, 4, -3) and parallel the line having Drs (1,-3, 7) is
$\begin{array}{l}\left|\begin{array}{ccc}\mathrm{x}-1& \mathrm{y}-2& \mathrm{z}+5\\ 2-1& 4+2& -3+5\\ 1& -3& 7\end{array}\right|=0\Rightarrow \left|\begin{array}{ccc}\mathrm{x}-1& \mathrm{y}-2& \mathrm{z}+5\\ 1& 2& 2\\ 1& -3& 7\end{array}\right|=0\\ \Rightarrow 4\mathrm{x}-\mathrm{y}-\mathrm{z}-7=0\end{array}$
Let $(\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma })$ be the image of the point (-1, 3, 4) in the plane Then 
$\begin{array}{l}\frac{\mathrm{\alpha }+1}{4}=\frac{\mathrm{\beta }-3}{-1}=\frac{\mathrm{\gamma }-4}{-1}=\frac{-2(-4-3-4-7)}{4^2+(-1{)}^2+(-1{)}^2}=2\\ \Rightarrow \mathrm{\alpha }=7,\mathrm{\beta }=1,\mathrm{\gamma }=2\Rightarrow \mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }=10\end{array}$