Let the system of linear equations \(-x+2 y-9 z=7\) \(-x+3 y+7 z=9\) \(-2 x+y+5 z=8\) \(-3 x+y+13 z=\lambda\) has a unique solution \(\mathrm{x}=\alpha, \mathrm{y}=\beta, \mathrm{z}=\gamma\). Then the distance of the point ( \(\alpha, \beta, \gamma\) ) from the plane \(2 \mathrm{x}-2 \mathrm{y}+\mathrm{z}=\lambda\) is

JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Matrices

Let the system of linear equations

- x + 2 y - 9 z = 7

- x + 3 y + 7 z = 9

- 2 x + y + 5 z = 8

- 3 x + y + 13 z = λ

has a unique solution

x = α, y = β, z = γ

. Then the distance of the point (

α, β, γ

) from the plane

2 x - 2 y + z = λ

Answer: (D)

Solve the first three equations. Subtracting the first from the second gives

y + 16 z = 2

; eliminating

x

between the first and third gives

3 y - 23 z = 6

. These yield

z = 0, y = 2

, and then

x = - 3

, so

(α, β, γ) = (- 3, 2, 0)

. The fourth equation gives

- 3 (- 3) + 2 + 13 (0) = λ

, so

λ = 11

. The distance from

(- 3, 2, 0)

to the plane

2 x - 2 y + z = 11

\frac{∣2 ( - 3 ) - 2 ( 2 ) + 0 - 11∣}{4 + 4 + 1} = \frac{21}{3} = 7

JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Single correct option