Let the image of the point $\mathrm{P}(1,2,6)$ in the plane passing through the points $\mathrm{A}(1,2,0), \mathrm{B}(1,4,1)$ and $\mathrm{C}(0,5,1)$ be $\mathrm{Q}(\alpha, \beta, \gamma)$. Then $\left(\alpha^{2}+\beta^{2}+\gamma^{2}\right)$ is equal to :

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Three Dimensional Geometry

Let the image of the point

P (1, 2, 6)

in the plane passing through the points

A (1, 2, 0), B (1, 4, 1)

and

C (0, 5, 1)

Q (α, β, γ)

. Then

(α^{2} + β^{2} + γ^{2})

is equal to :

(A)65✓ Correct
(B)70
(C)76
(D)62

Answer: (A)

Solution

The plane through

A (1, 2, 0), B (1, 4, 1), C (0, 5, 1)

has normal

A B \times A C = (- 1, - 1, 2)

, so its equation is

x + y - 2 z - 3 = 0

. At

P (1, 2, 6)

the expression

x + y - 2 z - 3

equals

- 12

, so the image is

Q = P - \frac{2 ( - 12 )}{6} (1, 1, - 2) = P + 4 (1, 1, - 2) = (5, 6, - 2)

. Therefore

α^{2} + β^{2} + γ^{2} = 25 + 36 + 4 = 65

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Single correct option

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