The mirror image of the point (1, 2, 3) in a | vedclass

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(D) Let the point be $P(1, 2, 3)$ and its image be $P'\left(-\frac{7}{3}, -\frac{4}{3}, -\frac{1}{3}\right)$.The midpoint $M$ of $PP'$ lies on the pla

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$(1, -1, 1)$

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(D) Let the point be $P(1, 2, 3)$ and its image be $P'\left(-\frac{7}{3}, -\frac{4}{3}, -\frac{1}{3}\right)$.The midpoint $M$ of $PP'$ lies on the plane:$M = \left(\frac{1 - \frac{7}{3}}{2}, \frac{2 - \frac{4}{3}}{2}, \frac{3 - \frac{1}{3}}{2}\right) = \left(-\frac{2}{3}, \frac{1}{3}, \frac{4}{3}\right)$.The normal vector $\vec{n}$ to the plane is given by the vector $\vec{PP'}$:$\vec{n} = \vec{PP'} = \left(-\frac{7}{3} - 1, -\frac{4}{3} - 2, -\frac{1}{3} - 3\right) = \left(-\frac{10}{3}, -\frac{10}{3}, -\frac{10}{3}\right)$.We can take the normal vector as $\vec{n} = (1, 1, 1)$.The equation of the plane is $1(x - (-\frac{2}{3})) + 1(y - \frac{1}{3}) + 1(z - \frac{4}{3}) = 0$.$x + \frac{2}{3} + y - \frac{1}{3} + z - \frac{4}{3} = 0 \implies x + y + z - 1 = 0 \implies x + y + z = 1$.Checking the options:For $(1, -1, 1)$,$1 + (-1) + 1 = 1$. Thus,the point $(1, -1, 1)$ lies on the plane.

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

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