JEE Mathematics 2023 Question with Solution

Given: The points $(1, \lambda, \frac{1}{2})$ and $(-2, 0, 1)$ are at equal distance from the plane $2x + 3y - 6z + 7 = 0$.

Find: The distance of the point $(1 - \lambda_2, \lambda_2, \lambda_1)$ from the line $\frac{x-5}{1} = \frac{y-1}{2} = \frac{z+7}{2}$.

Use the distance formula of a point $(x_1, y_1, z_1)$ from the plane $ax + by + cz + d = 0$:

From the given working, equating the distances of $(1, \lambda, \frac{1}{2})$ and $(-2, 0, 1)$ from the plane gives

Solving gives $\lambda_1 = 3$ and $\lambda_2 = 2$.

Hence the required point becomes

The line is parameterized, and the distance formula between a point and a line is then used.

Final Answer: The distance is $9$.