JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Three Dimensional Geometry

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:

(A)12
(B)7✓ Correct
(C)5
(D)17

Answer: (B)

Solution

The plane contains A

(7, 0, 6)

and B

(3, 4, 2)

, so

A B = - 4 \hat{i} + 4 \hat{j} - 4 \hat{k}

, and it is perpendicular to the plane

2 x - 5 y = 15

, whose normal is

2 \hat{i} - 5 \hat{j}

. The normal to the required plane is

A B \times (2 \hat{i} - 5 \hat{j}) = 5 \hat{i} + 2 \hat{j} - 3 \hat{k}

. Through A the plane is

5 (x - 7) + 2 y - 3 (z - 6) = 0

, i.e.

5 x + 2 y - 3 z = 17

. Putting the point

(2, α, β)

into it,

10 + 2 α - 3 β = 17

, so

2 α - 3 β = 7

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option

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