If the point (2, α, β) lies on the plane which passes through the points (3,4, 2) and (7, 0, 6) and is perpendicular to the plane 2x - 5y =15, then 2 α-3β is equal to

Question

If the point (2, α, β) lies on the plane which passes through the points (3,4, 2) and (7, 0, 6) and is perpendicular to the plane 2x - 5y =15, then 2 α-3β is equal to (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Let the normal to the required plane is ,

n

then

n = ∣ ∣ \hat{i} 42 \hat{j} - 4 - 5 \hat{k} 40 ∣ ∣ = 20 \hat{i} + 8 \hat{j} - 12 \hat{j}

∴

Equation of the plane

(x - 3) \times 20 + (y - 4) \times 8 + (z - 2) \times (- 12) = 0

5 x - 15 + 2 y - 8 - 3 z + 6 = 0

5 x + 2 y - 3 z - 17 = 0 \dots (1)

Since, equation of plane

(1)

passes through

(2, α, β)

,
then

10 + 2 α - 3 β - 17 = 0

\Rightarrow 2 α - 3 β = 7

Q. If the point (2,α,β) lies on the plane which passes through the points (3,4,2) and (7,0,6) and is perpendicular to the plane 2x−5y=15, then 2α−3β is equal to

Q. If the point $(2, α, β)$ lies on the plane which passes through the points $(3, 4, 2)$ and $(7, 0, 6)$ and is perpendicular to the plane $2 x - 5 y = 15$ , then $2 α - 3 β$ is equal to