Question

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 1

Correct Answer: 137

The correct answer is 137
\(\frac{x-a}{3} = \frac{y-b}{-4}=\frac{z-c}{12} \)
\(= \frac{-2(3a-4b+12c+19)}{3^2+(-4)^2+12^2}\)
\(\frac{x-a}{3} = \frac{y-b}{-4}=\frac{z-c}{12}\)
\(= \frac{-6a+8b-24c-38}{169}\)
(x,y,z) ≡ (a-6, β, λ)
\(\frac{(a-b)-a}{3} = \frac{β-b}{-4} = \frac{γ-c}{12} = \frac{-6a+8b-24c-38}{169}\)
\(\frac{β-b}{-4} = -2\)
⇒ β = 8+ b
\(\frac{γ-c}{12} = -2\)
⇒ Y = -24+c
\(\frac{-6a+8b-24c-38}{169} = -2\)
⇒ 3a – 4b + 12c = 150….(1)
a + b + c = 5
⇒ 3a + 3b + 3c = 15….(2)
Applying (1) – (2), we get;
-7b + 9c = 135
7b – 9c = -135
7β – 9γ = 7(8 + b) – 9 (–24 + c)
= 56 + 216 + 7b – 9c.
= 56 + 216 – 135 = 137.