Question

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_____.

Answer 1

$M =\left(a-3, \frac{\beta+b}{2}, \frac{\gamma+c}{2}\right)$
Since M lies on $3 x+4 y+12 z+19=0$
$\Rightarrow 6 a-4 b+12 c-4 \beta+12 \gamma+20=0$
Since PP' is parallel to normal of the plane then
$\frac{6}{3}=\frac{b-\beta}{-4}=\frac{c-\gamma}{12}$
$\Rightarrow \beta=b+8, \gamma=c-24$
$a+b+c=5 \Rightarrow a+\beta-8+\gamma+24=5$
$\Rightarrow a=-\beta-\gamma-11$
Now putting these values in (1) we get
$6(-\beta-\gamma-11)-4(\beta-8)+12(\gamma+24)-4 \beta+12 \gamma+20=0$
$\Rightarrow 7 \beta-9 \gamma=170-33=137$