If the system of linear equations $8 x+y+4 z=-2$ $x+y+z=0$ $\lambda \mathrm{x}-3 \mathrm{y}=\mu$ has infinitely many solutions, then the distance of the point $\left(\lambda, \mu,-\frac{1}{2}\right)$ from the plane $8 x+y+4 z+ 2=0$ is :

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Matrices

If the system of linear equations

8 x + y + 4 z = - 2

x + y + z = 0

λ x - 3 y = μ

has infinitely many solutions, then the distance of the point

(λ, μ, - \frac{1}{2})

from the plane

8 x + y + 4 z + 2 = 0

is :

Answer: (D)

For infinitely many solutions, the determinant of coefficients must be zero:

D = 81 λ 11 - 3 410 = 0

Expanding:

8 (0 + 3) - 1 (0 - λ) + 4 (- 3 - λ) = 0

, giving

24 + λ - 12 - 4 λ = 0

, so

λ = 4

.

For consistency, we also need

D_{1} = D_{2} = D_{3} = 0

. Checking

D_{1}

with the third equation: substituting

λ = 4

and solving gives

μ = - 2

.

The point is

(4, - 2, - 1/2)

. Distance from plane

8 x + y + 4 z + 2 = 0

d = \frac{∣8 ( 4 ) + ( - 2 ) + 4 ( - 1/2 ) + 2∣}{64 + 1 + 16} = \frac{∣32 - 2 - 2 + 2∣}{9} = \frac{30}{9} = \frac{10}{3}

JEE Main · 2022 · 26 Jul 2022 · Shift 1 · Single correct option