UPSC NDA I 2022 Mathematics Solved Paper

Concept:The given plane equation $6x + ky + 3z - 12 = 0$ is converted to intercept form. The sphere passes through the origin and the three intercept points on the axes. Substituting these points into the general sphere equation gives the value of $k$.Explanation:Write the plane in intercept form: $6x + ky + 3z = 12$ becomes $\frac{x}{2} + \frac{y}{12/k} + \frac{z}{4} = 1$.So the intercepts are $A(2,0,0)$, $B(0,\frac{12}{k},0)$, $C(0,0,4)$.Let the sphere be $x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$.Since it passes through the origin $(0,0,0)$, we get $d = 0$.Substitute $A(2,0,0)$: $4 + 2u \cdot 2 = 0 \Rightarrow 4 + 4u = 0 \Rightarrow u = -1$.Substitute $B(0,\frac{12}{k},0)$: $\frac{144}{k^2} + 2v \cdot \frac{12}{k} = 0 \Rightarrow \frac{144}{k^2} + \frac{24v}{k} = 0$ → multiply by $k^2$: $144 + 24vk = 0 \Rightarrow v = -\frac{6}{k}$.Substitute $C(0,0,4)$: $16 + 2w \cdot 4 = 0 \Rightarrow 16 + 8w = 0 \Rightarrow w = -2$.The sphere equation becomes: $x^2 + y^2 + z^2 - 2x - \frac{12}{k}y - 4z = 0$.Comparing with the given equation $x^2 + y^2 + z^2 - 2x - 3y - 4z = 0$, the coefficients of $y$ must match: $-\frac{12}{k} = -3$ ⇒ $\frac{12}{k} = 3$ ⇒ $k = 4$.Answer:$k = 4$ (Option B).