$\begin{array}{l}\text{Let }O\text{\hspace{0.17em} be the origin. Let }\overline{OP}=xi+yj-k,\overline{OQ}=-i+2j+3xk,\text{ and }\overline{OR}=3i+zj-7kx,y,z\in R,x>0\\ \text{be such that }\left|PQ\right|\text{=}\sqrt{20}\text{, }\overline{OP}\text{\hspace{0.17em} is orthogonal to }\overline{OQ},P,Q,R\text{\hspace{0.33em} are collinear\hspace{0.17em} then the value of }{x}^2+y^2+z^2\end{array}$

$\begin{array}{l}\overline{OP}\perp \overline{OQ}\Rightarrow -x+2y-3x=0\Rightarrow y=2x\dots \\ \left|\overline{PQ}\right|=\sqrt{2}0\Rightarrow (x+1{)}^2+(2-y{)}^2+(3x+1{)}^2=20\\ \Rightarrow 14x^2-14=0\Rightarrow x=\pm 1,x>0 x=1, y=2\\ \mathrm{OP},\mathrm{OQ},\mathrm{OR}\text{ are collinear }\end{array}$

$\begin{array}{c}\left|\begin{array}{ccc}x& y& -1\\ -1& 2& 3x\\ 3& z& -7\end{array}\right|=0\\ x\left(-14-3xz\right)-y\left(7-9x\right)-1\left(-z-6\right)=0\\ -14-3z-2\left(-2\right)+z+6=0\\ -14-3z+4+z+6=0\\ -2z=4\\ z=-2\end{array}$

$\text{Therefore, }{x}^2+y^2+z^2=1+4+4=9$