Question

Let the plane ax + by + cz = d pass through (2, 3, -5) and is perpendicular to the planes 2x + y - 5z = 10 and 3x + 5y - 7z = 12. If a, b, c, d are integers d > 0 and gcd (|a|, |b|, |c|, d) = 1, then the value of a + 7b + c + 20d is equal to :

• A. 18
• B. 20
• C. 24
• D. 22

Answer 1

Answer: (D)

22

Answer 2

The correct answer is (D) : 22
Equation of plane through point (2, 3, –5) and perpendicular to planes 2x + y – 5z = 10 and 3x + 5y – 7z = 12 is
$\begin{vmatrix} x-2 & y-3 & z+5 \\\ 2 & 1 & -5 \\\ 3 & 5 & -7 \\\ \end{vmatrix} = 0$
∴ Equation of plane is (x – 2) (– 7 + 25) – (y – 3) (– 14 + 15) + (z + 5) · 7 = 0
∴ 18x – y + 7z + 2 = 0
⇒ 18x – y + 7z = – 2
∴ – 18x + y – 7z = 2
On comparing with ax + by + cz = d where d > 0 is
a = – 18, b = 1, c = – 7, d = 2
Therefore a + 7b + c + 20d = 22