A variable plane passes through a fixed point $(a, b, c)$ and cuts the axes in $\mathrm{A}, \mathrm{B}$ and C respectively. The locus of the centre of the sphere $\mathrm{OABC}, \mathrm{O}$ being the origin, is

• A

  $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}+\frac{\mathrm{z}}{\mathrm{c}}=1$
• B

  $\frac{\mathrm{a}}{\mathrm{x}}+\frac{\mathrm{b}}{\mathrm{y}}+\frac{\mathrm{c}}{\mathrm{z}}=1$
• Correct answer
  C

  $\frac{\mathrm{a}}{\mathrm{x}}+\frac{\mathrm{b}}{\mathrm{y}}+\frac{\mathrm{c}}{\mathrm{z}}=2$
• D

  $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}+\frac{\mathrm{z}}{\mathrm{c}}=2$

Solution

Equation of plane passing through points $(p, 0,0),(0, q, 0)$ and $(0,0, r)$ is $\frac{x}{p}+\frac{y}{q}+\frac{z}{r}=1$ Given that this plane passes through $(a, b, c)$. $\therefore \frac{\mathrm{a}}{\mathrm{p}}+\frac{\mathrm{b}}{\mathrm{q}}+\frac{\mathrm{c}}{\mathrm{r}}=1$ Equation of sphere is $x^2+y^2+z^2-p x-q y-r z=0$. Centre of the sphere $=(1, m, n)=\left(\frac{\mathrm{p}}{2}, \frac{\mathrm{q}}{2}, \frac{\mathrm{r}}{2}\right)$ $\Rightarrow \mathrm{p}=2 \ell, \mathrm{q}=2 \mathrm{~m}, \mathrm{r}=2 \mathrm{n}$ $\therefore$ locus of the centre $\Rightarrow \frac{\mathrm{a}}{\mathrm{x}}+\frac{\mathrm{b}}{\mathrm{y}}+\frac{\mathrm{c}}{\mathrm{z}}=2$.

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