UPSC NDA I 2016 Mathematics Solved Paper

Concept:The plane $P$ is a member of the family of planes passing through the line of intersection of the two given planes.Explanation:Consider the two planes: $2x - y + 3z - 2 = 0$ and $x + y - z - 1 = 0$.Any plane through their line of intersection can be written as: $(2x - y + 3z - 2) + \lambda (x + y - z - 1) = 0$.This simplifies to: $(2 + \lambda)x + (-1 + \lambda)y + (3 - \lambda)z + (-2 - \lambda) = 0$.The plane passes through the point $(1, 0, 1)$. Substitute $x = 1$, $y = 0$, $z = 1$:$(2 + \lambda)(1) + (-1 + \lambda)(0) + (3 - \lambda)(1) + (-2 - \lambda) = 0$.This gives: $2 + \lambda + 0 + 3 - \lambda - 2 - \lambda = 0$, i.e., $3 - \lambda = 0$.Hence $\lambda = 3$.Substitute $\lambda = 3$ back into the combined equation: $(2+3)x + (-1+3)y + (3-3)z + (-2-3) = 0$.Thus $5x + 2y + 0z - 5 = 0$, or $5x + 2y - 5 = 0$.Answer:$5x + 2y - 5 = 0$ (Option B).