Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

$\begin{array}{rl}& A=\left(\frac{5r-1}{r+1},\frac{5r-1}{r+1},\frac{10r+2}{r+1}\right)\\ & (\overrightarrow{OQ}\cdot \overrightarrow{OA})-\frac{1}{5}|\overrightarrow{OP}\times \overrightarrow{OA}{|}^2=10\\ & \overrightarrow{OQ}=5\hat{i}+5\hat{j}+10\hat{k}\\ & \overrightarrow{OA}=\frac{5r-1}{r+1}\hat{i}+\frac{5r-1}{r+1}\hat{j}+\frac{10r+2}{r+1}\hat{k}\\ & \overrightarrow{OP}=-\hat{i}-\hat{j}+2\hat{k}\\ & \overrightarrow{OP}\times \overrightarrow{OA}=\frac{1}{r+1}|\frac{5r-1}{5}5r-110r+2\\ & =\frac{1}{r+1}(\hat{i}(20r)-\hat{j}(20r))\\ & =5\left(\frac{5r-1}{r+1}\right)+5\left(\frac{5r-1}{r+1}\right)+10\left(\frac{10r+2}{r+1}\right)\\ & -\frac{1}{5}\left(\frac{2\times 400r^2}{(r+1{)}^2}\right)=10\\ & \frac{150r+10}{r+1}-\frac{1}{5}\left(\frac{2\times 400r^2}{(r+1{)}^2}\right)=10\\ & (150r+10)(r+1)-160r^2=10(r+1{)}^2\\ & (15r+1)(r+1)-16r^2=(r+1{)}^2\\ & 15r^2+16r+1-16r^2=r^2+2r+1\\ & -2r^2+14r=0\end{array}$
$r=0,7$