Let A, B, C be three points on overline{OX}, | vedclass

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(C) Let the coordinates of point $P$ be $(u, v, w)$.Since $P$ is equidistant from $O(0, 0, 0), A(3, 0, 0), B(0, 6, 0),$ and $C(0, 0, 9)$,we have $PO^2

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$\left(\frac{9}{5}, \frac{21}{5}, \frac{33}{5}\right)$

Vedclass · Answer

(C) Let the coordinates of point $P$ be $(u, v, w)$.Since $P$ is equidistant from $O(0, 0, 0), A(3, 0, 0), B(0, 6, 0),$ and $C(0, 0, 9)$,we have $PO^2 = PA^2 = PB^2 = PC^2$.$PO^2 = u^2 + v^2 + w^2$.$PA^2 = (u-3)^2 + v^2 + w^2 = u^2 - 6u + 9 + v^2 + w^2$.Equating $PO^2 = PA^2$,we get $u^2 = u^2 - 6u + 9$ $\Rightarrow 6u = 9$ $\Rightarrow u = \frac{3}{2}$.$PB^2 = u^2 + (v-6)^2 + w^2 = u^2 + v^2 - 12v + 36 + w^2$.Equating $PO^2 = PB^2$,we get $v^2 = v^2 - 12v + 36$ $\Rightarrow 12v = 36$ $\Rightarrow v = 3$.$PC^2 = u^2 + v^2 + (w-9)^2 = u^2 + v^2 + w^2 - 18w + 81$.Equating $PO^2 = PC^2$,we get $w^2 = w^2 - 18w + 81$ $\Rightarrow 18w = 81$ $\Rightarrow w = \frac{81}{18} = \frac{9}{2}$.Thus,$P = \left(\frac{3}{2}, 3, \frac{9}{2}\right)$.Given $Q = (2, 5, 8)$,point $R$ divides $PQ$ in the ratio $3:2$. Using the section formula $\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n}\right)$:$R = \left(\frac{3(2) + 2(\frac{3}{2})}{3+2}, \frac{3(5) + 2(3)}{3+2}, \frac{3(8) + 2(\frac{9}{2})}{3+2}\right)$$R = \left(\frac{6+3}{5}, \frac{15+6}{5}, \frac{24+9}{5}\right) = \left(\frac{9}{5}, \frac{21}{5}, \frac{33}{5}\right)$.

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