Answer
$P = \left( {\frac{2}{3}, - \frac{1}{3},\frac{2}{3}} \right)$
Work Step by Step
By Theorem 1, the plane $x+y+z=1$ has normal vector ${\bf{n}} = \left( {1,1,1} \right)$.
By Eq. (7) of Exercise 71, the point $P$ on the plane closest to $Q$ is determined by the equation:
$\overrightarrow {OP} = \overrightarrow {OQ} + \left( {\frac{{d - \overrightarrow {OQ} \cdot{\bf{n}}}}{{{\bf{n}}\cdot{\bf{n}}}}} \right){\bf{n}}$
We have $\overrightarrow {OQ} = Q - O = \left( {2,1,2} \right)$. Substituting the corresponding values in Eq. (7) gives
$\overrightarrow {OP} = \left( {2,1,2} \right) + \left( {\frac{{1 - \left( {2,1,2} \right)\cdot\left( {1,1,1} \right)}}{{\left( {1,1,1} \right)\cdot\left( {1,1,1} \right)}}} \right)\left( {1,1,1} \right)$
$\overrightarrow {OP} = \left( {2,1,2} \right) + \left( {\frac{{1 - 5}}{3}} \right)\left( {1,1,1} \right)$
$\overrightarrow {OP} = \left( {2,1,2} \right) - \left( {\frac{4}{3},\frac{4}{3},\frac{4}{3}} \right) = \left( {\frac{2}{3}, - \frac{1}{3},\frac{2}{3}} \right)$
Thus, the point $P = \left( {\frac{2}{3}, - \frac{1}{3},\frac{2}{3}} \right)$ is nearest to $Q = \left( {2,1,2} \right)$ on the plane $x+y+z=1$.
