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

$\begin{array}{rl}& \mathrm{A}(3,\alpha ,3)\&B(-3,-7,\beta )\\ & \overrightarrow{\mathrm{BA}}=6\hat{\mathrm{i}}+(\alpha +7)\hat{\mathrm{j}}+(3-\beta )\hat{\mathrm{k}}\end{array}$
$\overrightarrow{\mathrm{p}}\times \overrightarrow{\mathrm{q}}=\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 3& -1& 1\\ -3& 2& 4\end{array}\right|$
$\frac{|\overrightarrow{\mathrm{BA}}\cdot (\overrightarrow{\mathrm{p}}\times \overrightarrow{\mathrm{q}})|}{|\overrightarrow{\mathrm{p}}\times \overrightarrow{\mathrm{q}}|}=3\sqrt{30}$
$36+15(\alpha +7)-3(3-\beta )=(3\sqrt{30}{)}^2$
$\begin{array}{rl}& 36+15\alpha +105-9+3\beta =270\\ & 15\alpha +3\beta =138\\ & 5\alpha +\beta =46\end{array}$