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

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Three Dimensional Geometry

Let the plane

ax + by + cz = d

pass through

(2, 3, - 5)

and is perpendicular to the planes

2 x + y - 5 z = 10

and

3 x + 5 y - 7 z = 12

. If

a, b, c, d

are integers

d > 0

and

gcd (∣ a ∣, ∣ b ∣, ∣ c ∣, d) = 1

, then the value of

a + 7 b + c + 20 d

is equal to

Answer: (D)

The normal to the required plane is perpendicular to both given planes, so it is parallel to their cross product:

\hat{i} 23 \hat{j} 15 \hat{k} - 5 - 7 = 18 \hat{i} - \hat{j} + 7 \hat{k}

. The plane equation is

18 x - y + 7 z = d

. Passing through

(2, 3, - 5)

36 - 3 - 35 = d = - 2

. To satisfy

d > 0

and

g cd = 1

, we write

- 18 x + y - 7 z = 2

, giving

a = - 18, b = 1, c = - 7, d = 2

. Therefore,

a + 7 b + c + 20 d = - 18 + 7 - 7 + 40 = 22

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Single correct option