A(1, 2, 3), B(2, 3, 1), and C(3, 1, 2) are th | vedclass

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(C) The coordinates of point $P$ dividing $AB$ in the ratio $m:n = 1:2$ are given by the section formula: $P = \left( \frac{mx_2 + nx_1}{m+n}, \frac{m

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$\frac{2}{3} \sqrt{78}$

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(C) The coordinates of point $P$ dividing $AB$ in the ratio $m:n = 1:2$ are given by the section formula: $P = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n} \right)$ $P = \left( \frac{1(2) + 2(1)}{1+2}, \frac{1(3) + 2(2)}{1+2}, \frac{1(1) + 2(3)}{1+2} \right) = \left( \frac{4}{3}, \frac{7}{3}, \frac{7}{3} \right)$ The coordinates of point $Q$ dividing $BC$ in the ratio $m:n = -2:3$ are: $Q = \left( \frac{-2(3) + 3(2)}{-2+3}, \frac{-2(1) + 3(3)}{-2+3}, \frac{-2(2) + 3(1)}{-2+3} \right) = \left( \frac{0}{1}, \frac{7}{1}, \frac{-1}{1} \right) = (0, 7, -1)$ The distance $PQ$ is given by $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$: $PQ = \sqrt{\left(0 - \frac{4}{3}\right)^2 + \left(7 - \frac{7}{3}\right)^2 + \left(-1 - \frac{7}{3}\right)^2}$ $PQ = \sqrt{\left(-\frac{4}{3}\right)^2 + \left(\frac{14}{3}\right)^2 + \left(-\frac{10}{3}\right)^2}$ $PQ = \sqrt{\frac{16}{9} + \frac{196}{9} + \frac{100}{9}} = \sqrt{\frac{312}{9}} = \frac{\sqrt{312}}{3} = \frac{2\sqrt{78}}{3}$

$A(1, 2, 3)$, $B(2, 3, 1)$, and $C(3, 1, 2)$ are three points. If the point $P$ divides $AB$ in the ratio $1:2$ and the point $Q$ divides $BC$ in the ratio $-2:3$, then the distance between $P$ and $Q$ is

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