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$A(1,2,3), B(2,3,1)$ and $C(3,1,2)$ are three points. If the point $\mathrm{P}$ divides $\mathrm{AB}$ in the ratio $1: 2$ and the point $\mathrm{Q}$ divides $\mathrm{BC}$ in the ratio $-2: 3$, then the distance between $P$ and $Q$ is
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Verified Answer
The correct answer is:
$\frac{2}{3} \sqrt{78}$
$$
\begin{aligned}
& x=\frac{2+2}{3}, y=\frac{3+4}{3}, z=\frac{1+6}{3} \\
& P(x, y, z)=\left(\frac{4}{3}, \frac{7}{3}, \frac{7}{3}\right) \\
& \text {. } \begin{array}{ll}
& -2,1 \\
B & Q
\end{array} \\
& (2,3,1) \quad(\alpha, \beta, \gamma) \quad(3,1,2) \\
& \alpha=\frac{-6+6}{1}, \beta=\frac{-2+9}{1}, \gamma=\frac{-4+3}{1} \\
& Q(\alpha, \beta, \gamma)=(0,7,-1) \\
& \therefore P Q=\sqrt{\left(\frac{4}{3}-0\right)^2+\left(\frac{7}{3}-7\right)^2+\left(\frac{7}{3}+1\right)^2} \\
& =\sqrt{\frac{16}{9}+\frac{196}{9}+\frac{100}{9}}=\frac{\sqrt{312}}{3}=\frac{2}{3} \sqrt{78} . \\
&
\end{aligned}
$$
\begin{aligned}
& x=\frac{2+2}{3}, y=\frac{3+4}{3}, z=\frac{1+6}{3} \\
& P(x, y, z)=\left(\frac{4}{3}, \frac{7}{3}, \frac{7}{3}\right) \\
& \text {. } \begin{array}{ll}
& -2,1 \\
B & Q
\end{array} \\
& (2,3,1) \quad(\alpha, \beta, \gamma) \quad(3,1,2) \\
& \alpha=\frac{-6+6}{1}, \beta=\frac{-2+9}{1}, \gamma=\frac{-4+3}{1} \\
& Q(\alpha, \beta, \gamma)=(0,7,-1) \\
& \therefore P Q=\sqrt{\left(\frac{4}{3}-0\right)^2+\left(\frac{7}{3}-7\right)^2+\left(\frac{7}{3}+1\right)^2} \\
& =\sqrt{\frac{16}{9}+\frac{196}{9}+\frac{100}{9}}=\frac{\sqrt{312}}{3}=\frac{2}{3} \sqrt{78} . \\
&
\end{aligned}
$$
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