UPSC NDA II 2025 Mathematics Paper

Concept:The required plane passes through the line of intersection of two given planes. It must be parallel to a given line, so its normal vector is perpendicular to the direction vector of that line.Explanation:Start with the family of planes: $(6x+4y-5z-2)+\lambda(x-2y+3z)=0$.Expand: $(6+\lambda)x+(4-2\lambda)y+(-5+3\lambda)z-2=0$.Normal vector $\vec{n}=(6+\lambda,\;4-2\lambda,\;-5+3\lambda)$.Direction vector of the line: $\vec{d}=(1,3,2)$.Parallel condition: $\vec{n}\cdot\vec{d}=0$.Compute: $(6+\lambda)\cdot1+(4-2\lambda)\cdot3+(-5+3\lambda)\cdot2=0$.Simplify: $6+\lambda+12-6\lambda-10+6\lambda=0 \;\Rightarrow\; \lambda+8=0 \;\Rightarrow\; \lambda=-8$.Substitute $\lambda=-8$: $(6-8)x+(4+16)y+(-5-24)z-2=0$ gives $-2x+20y-29z-2=0$.Multiply by $-1$ to get positive coefficients: $2x-20y+29z+2=0$.Answer:Option A: $2x-20y+29z+2=0$