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

$\mathrm{Equation} \mathrm{of} \mathrm{given} \mathrm{plane} \mathrm{is} 3x-4y+12z+19=0$

Direction ration of the line  $LL\text{'}$   is  $(6,6-\beta ,c-\gamma )$  Midpoint of line  $LL\text{'}$  is 

 $M=(a-3,\frac{\beta +b}{2},\frac{\gamma +c}{2})$
Here M lies on  $3x-4y+12z+19=0$
Satisfy point M in the above plane.
 $\Rightarrow 3a-2\beta -2b+6\gamma +6c+10=0$
Here, line  $LL"$ is parallel to normal of the plane then
 $\frac{6}{3}=\frac{b-\beta }{-4}=\frac{c-\gamma }{12} \Rightarrow b=\beta -8,c=\gamma +24$
 Put the value of b & c in the given equation.
 $$\begin{gathered} a+b+c=5\Rightarrow \alpha +\beta -8+\gamma +24=5 \\ \Rightarrow \alpha =\beta -\gamma -11 \end{gathered}$$
 Put these values in (i), we get
 $$\begin{gathered} 3(-\beta -\gamma -11)-2\beta -2(\beta -8)+6\gamma +6(\gamma +24)+10=0 \\ -7\beta +9\gamma +137=0\Rightarrow -7\beta +9\gamma +137=0 \\ -7\beta +9\gamma =-137 \end{gathered}$$