Q243 / 270 · NDA 2022 (Phase 1) · Mathematics

The plane $6 x+k y+3 z-12=0$ where meets the coordinate axes at $A, B$ and $C$ respectively. The equation of the sphere passing through the origin and $A, B, C$ is $x^2+y^2+z^2-2 x-$ $3 y-4 z=0$. What is the equation of the line through the origin and the centre of the sphere?

  • A
    $x=y=z$
  • B
    $2 x=3 y=4 z$
  • C
    $6 x=3 y=4 z$
  • Correct answer
    D
    $6 x=4 y=3 z$

Solution

Equation of line passing through $(0,0,0)$ and $\left(1, \frac{3}{2}, 2\right)$ is $\frac{x-0}{1}=\frac{y-0}{\frac{3}{2}}=\frac{z-0}{2}$ $6 x=4 y=3 z$ For question 94 and 95 Since point $(2,3,-6)$ lies plane $\begin{aligned} & \quad \frac{2 \mathrm{x}}{\mathrm{k}}+\frac{2 \mathrm{y}}{3}+\frac{\mathrm{z}}{3}=2 \\ & \therefore \quad \frac{4}{\mathrm{k}}+\frac{6}{3}-\frac{6}{3}=2 \\ & \Rightarrow \frac{4}{\mathrm{k}}=2 \Rightarrow \mathrm{k}=2\end{aligned}$ So, equation of plane is $\begin{aligned} & x+\frac{2 y}{3}+\frac{z}{3}=2 \\ & \Rightarrow 3 x+2 y+z=6\end{aligned}$

Hence, the correct answer is option: 'D'.

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