3D Geometry
Three Dimensional Geometry
star_batch_jee_advanced_2025
Grade 12
Question:
The coordinates of points, whose perpendicular distances from $yz, zx$ and $xy$-planes are in A.P., and whose distances from $x, y$ and $z$ axes are $\sqrt{13}, \sqrt{10}$ and $\sqrt{5}$ respectively is:
(1, 2, 3)
(-1, 2, 3)
(1, -2, 3)
(-1, -2, -3)
Step-by-Step Solution
Key Concept: Perpendicular distances to coordinate planes form an AP, and solving the resulting system of equations yields integer coordinates.
Let the perpendicular distances from point $P$ to the coordinate planes be $|a-d|$, $|a|$, and $|a+d|$ respectively, forming an arithmetic progression. Setting up three equations from the given conditions: $2a^2 + a^2 - 2ad = 5$, $2a^2 + 2d^2 = 10$, and $2a^2 + d^2 + 2ad = 13$. Solving equations (1) and (3) yields $4ad = 8$, so $ad = 2$. Substituting into equation (2): $\frac{4}{d^2} + d^2 = 5$, giving $d^2 = 1$ or $4$, thus $d = \pm 1, a = \pm 2$. The eight points are $(\pm 1, \pm 2, \pm 3)$.
Correct Answer: 1