The mirror image of the point $(1,2,3)$ in a plane is $\left(-\frac{7}{3},-\frac{4}{3},-\frac{1}{3}\right)$.…

The mirror image of the point $(1,2,3)$ in a plane is $\left(-\frac{7}{3},-\frac{4}{3},-\frac{1}{3}\right)$. Thus, the point lies on this plane.
  1. $(1,-1,1)$
  2. $(-1,-1,1)$
  3. $(1,1,1)$
  4. $(-1,-1,-1)$

Solution

$\mathrm{M}$ is the midpoint. $\therefore \quad \mathrm{M} \equiv\left(-\frac{2}{3}, \frac{1}{3}, \frac{4}{3}\right)$ D.r.s of $\mathrm{AB}$ are $\frac{-10}{3}, \frac{-10}{3}, \frac{-10}{3}$ i.e., $1,1,1$ Equation of plane is $\begin{aligned} & 1\left(x+\frac{2}{3}\right)+1\left(y-\frac{1}{3}\right)+1\left(z-\frac{4}{3}\right)=0 \\ & \Rightarrow x+y+z=1 \end{aligned}$ Option (A) satisfies this equation of the plane.

Asked in: MHT CET 2023 (11 May Shift 1)

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