A line from the origin meets the lines $\frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-1}{-2}=\frac{\mathrm{z}+1}{1}\text{ and }\frac{\mathrm{x}-\frac{8}{3}}{2}=\frac{\mathrm{y}+3}{-1}=\frac{\mathrm{z}-1}{1}$ at P and Q respectively. If length PQ=d, then the value of ${\mathrm{d}}^2$ is______

A point on the line$\frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}-1}{-2}=\frac{\mathrm{z}+1}{1}=s \mathrm{is} \mathrm{P}\left(\mathrm{s}+2,-2\mathrm{s}+1,\mathrm{s}-1\right)$

A point on the line$\frac{\mathrm{x}-\frac{8}{3}}{2}=\frac{\mathrm{y}+3}{-1}=\frac{\mathrm{z}-1}{1}=t \mathrm{is}$

$\mathrm{Q}=\left(\frac{8}{3}+2\mathrm{t}, -3-\mathrm{t}, 1+\mathrm{t}\right)$

O, P, Q are collinear, $\Rightarrow \frac{2+\mathrm{s}}{\frac{8}{3}+2\mathrm{t}}=\frac{1-2\mathrm{s}}{-3-\mathrm{t}}=\frac{-1+\mathrm{s}}{1+\mathrm{t}}$

Solving, $\mathrm{s}=3, \mathrm{t}=\frac{1}{3},$

$$\begin{gathered} \Rightarrow \mathrm{P}=(5,-5,2), \mathrm{Q}=\left(\frac{10}{3},-\frac{10}{3},\frac{4}{3}\right) \\ \Rightarrow P{Q}^2={\left(5-\frac{10}{3}\right)}^2+{\left(-5+\frac{10}{3}\right)}^2+{\left(2-\frac{4}{3}\right)}^2=\frac{25}{9}+\frac{25}{9}+\frac{4}{9}=6 \end{gathered}$$