JEE Mathematics 2023 Question with Solution

Given: The point is $P(1, 2, 6)$ and its image in the plane through $A(1, 2, 0)$, $B(1, 4, 1)$ and $C(0, 5, 1)$ is $Q(\alpha, \beta, \gamma)$.

Find: $\alpha^2 + \beta^2 + \gamma^2$.

From the solution, the equation of the plane is taken as

$$A(x - 1) + B(y - 2) + Cz = 0$$

Using point $B(1, 4, 1)$,

$$2B + C = 0$$

Using point $C(0, 5, 1)$,

$$-A + 3B + C = 0$$

Solving,

$$A = -2B, \quad C = -2B$$

So a normal vector is proportional to

$$(1, 1, -2)$$

Using the image formula written in the solution,

$$\frac{\alpha - 1}{1} = \frac{\beta - 2}{1} = \frac{\gamma - 6}{-2} = \frac{-2(1 + 2 - 12 - 3)}{6}$$

Hence,

$$\alpha = 5, \quad \beta = 6, \quad \gamma = -2$$

Now,

$$\alpha^2 + \beta^2 + \gamma^2 = 5^2 + 6^2 + (-2)^2 = 25 + 36 + 4 = 65$$

Therefore, the value of $\alpha^2 + \beta^2 + \gamma^2$ is $65$.

The extracted working gives final value $65$, which corresponds to option D. The provided solution labels the correct option as B, creating a discrepancy; the computed value is used as the answer authority.

Given: The reflected point coordinates obtained in the solution are $\alpha = 5$, $\beta = 6$, $\gamma = -2$.

Find: The required sum.

Substitute directly:

$$\alpha^2 + \beta^2 + \gamma^2 = 5^2 + 6^2 + (-2)^2$$ $$= 25 + 36 + 4 = 65$$

So the correct numerical value is $65$.

Among the listed options, $65$ is option D.