Let the mirror image of the point (a,b,c) with respect to the plane 3x−4y+12z+19=0 be (a−6,β,γ). If a+b+c=5, then 7β−9γ is equal to:
Answer: 137
Solutions · 1
Official solution
\[\frac{x-a}{3}=\frac{y-b}{-4}=\frac{z-c}{12}=\frac{-2(3a-4b+12c+19)}{169}.\]
Matching image coordinates: 3−6=−2 (consistent for the x-component), and this common ratio equals −2, giving
\[\beta=8+b,\qquad \gamma=-24+c,\qquad 3a-4b+12c=150.\]
Combined with a+b+c=5 (i.e. 3a+3b+3c=15), subtracting:
\[-7b+9c=135 \implies 7b-9c=-135.\]
\[7\beta-9\gamma=7(8+b)-9(-24+c)=56+216+7b-9c=272-135=137.\]