Question

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_____.

Answer 1

Answer: 137

$M=\left(a-3,\frac{\beta +b}{2},\frac{\gamma +c}{2}\right)$
Since M lies on $3x+4y+12z+19=0$
$\Rightarrow 6a-4b+12c-4\beta +12\gamma +20=0$
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
$6(-\beta -\gamma -11)-4(\beta -8)+12(\gamma +24)-4\beta +12\gamma +20=0$
$\Rightarrow 7\beta -9\gamma =170-33=137$