UPSC NDA II 2024 Mathematics Paper

Concept:The intersection point of a line given by two planes and a third plane must satisfy all three plane equations simultaneously.Explanation:The line $L$ is the intersection of two planes: $x + y + z + 4 = 0$ and $2x - y - z + 8 = 0$.The plane $P$ is $x + 2y + 3z + 1 = 0$.The intersection point must satisfy all three equations.Check each option by substituting the coordinates.Option A $(4,3,-3)$: $4+3-3+4 = 8 \neq 0$; fails the first equation.Option B $(4,-3,3)$: $4-3+3+4 = 8 \neq 0$; fails.Option C $(-4,-3,-3)$: $-4-3-3+4 = -6 \neq 0$; fails.Option D $(-4,-3,3)$: $-4-3+3+4 = 0$; satisfies the first equation of $L$.Next, $2(-4) - (-3) - 3 + 8 = -8+3-3+8 = 0$; satisfies the second equation of $L$.Finally, $-4 + 2(-3) + 3(3) + 1 = -4-6+9+1 = 0$; satisfies $P$.Thus only option D satisfies all three equations.Answer:D. $(-4, -3, 3)$