Let the mirror image of the point (a, b, c) with respect to the plane $3 x-4 y+12 z+19=0$ be $(a-6, \beta , \gamma )$. If $a+b+c=5$, then $7\beta -9\gamma$ is equal to______

Given that the image of $A\left(a,b,c\right)$ with respect to the plane $3 x-4 y+12 z+19=0$ be $B (a-6, \beta , \gamma )$

The mid point of $AB$ is  $M$ lies on the plane $3 x-4 y+12 z+19=0$

$\mathrm{M}=\left(a-3, \frac{\beta +b}{2}, \frac{\gamma +c}{2}\right)$

$\Rightarrow 6a-4b+12c-4\beta +12\gamma +20=0$      

Since $AB$ is parallel to normal of the plane then

$\frac{6}{3}=\frac{b-\beta }{-4}=\frac{c-\gamma }{12}\Rightarrow \beta =b+8, \gamma =c-24$

$\begin{array}{l}Given a+b+c=5 \\ \Rightarrow a+\beta -8+\gamma +24=5 \\ \Rightarrow a=-\beta -\gamma -11\end{array}$

Now putting these values in $3 x-4 y+12 z+19=0$ we get 

$\begin{array}{l}6(-\beta -\gamma -11)-4(\beta -8)+12(\gamma +24)-4\beta +12\gamma +20=0\\ \\ \Rightarrow 7\beta -9\gamma =170-33=137\end{array}$