3D Geometry
Three Dimensional Geometry
star_batch_jee_advanced_2025
Grade 12
Question:
If three planes $P_1: 2x + y + z - 1 = 0$, $P_2: x - y + z - 2 = 0$ and $P_3: ax - y + 3z - 5 = 0$ intersect each other at point $P$ on $XOY$ plane and at point $Q$ on $YOZ$ plane, where $O$ is the origin then identify the correct statement(s)
the value of $a$ is $4$
straight line perpendicular to plane $P_3$ and passing through $P$ is $\frac{x-1}{4} = \frac{y+1}{-1} = \frac{z}{3}$
the length of projection of $PQ$ on $x$-axis is $1$
centroid of the triangle $OPQ$ is $\left(\frac{1}{3}, -\frac{1}{2}, \frac{1}{2}\right)$
Step-by-Step Solution
Key Concept: When three planes intersect at a line, any two equations determine that line; the third confirms consistency.
Three planes meet at two points means they have infinitely many solutions. Setting up the system with planes $R: 2x + y + z = 1$, $P_2: x - y + z = 2$, and $P_3: 4x + y + 3z = 5$, we find the line of intersection by solving any two equations. On the $XOY$ plane (setting $z = 0$), we get point $P(1, -1, 0)$. On the $YOZ$ plane (setting $x = 0$), we get point $Q(0, - rac{1}{2}, rac{3}{2})$.
Correct Answer: Let me verify each statement systematically.
**Finding Point P (on XOY plane, z=0):**
From $P_1: 2x + y + 0 = 1$ → $2x + y = 1$
From $P_2: x - y + 0 = 2$ → $x - y =