JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Three Dimensional Geometry

Let the line

\frac{x - 3}{7} = \frac{y - 2}{- 1} = \frac{z - 3}{- 4}

intersect the plane containing the lines

\frac{x - 4}{1} = \frac{y + 1}{- 2} = \frac{z}{1}

and

4 a x - y + 5 z - 7 a = 0 = 2 x - 5 y - z - 3

a \in R

at the point

P (α, β, γ)

. Then the value of

α + β + γ

equals ____.

Answer: 12

Solution

The plane containing the two given lines has equation:

4 a x - y + 5 z - 7 a + λ (2 x - 5 y - z - 3) = 0

Since it contains point

(4, - 1, 0)

16 a + 1 - 7 a + λ (8 + 5 - 3) = 0

9 a + 1 + 10 λ = 0

... (1)

The direction vector

(1, - 2, 1)

of the line lies in the plane:

(4 a + 2 λ) (1) + (- 1 - 5 λ) (- 2) + (5 - λ) (1) = 0

4 a + 2 λ + 2 + 10 λ + 5 - λ = 0

4 a + 11 λ + 7 = 0

... (2)

Solving (1) and (2):

a = 1

λ = - 1

Plane equation:

x + 2 y + 3 z - 2 = 0

Point on the given line:

(7 t + 3, - t + 2, - 4 t + 3)

Substituting:

7 t + 3 - 2 t + 4 - 12 t + 9 - 2 = 0

- 7 t + 14 = 0

, so

t = 2

Point

P = (17, 0, - 5)

α + β + γ = 17 + 0 - 5 = 12

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Integer answer