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

$\begin{array}{rl}& \overrightarrow{\mathrm{p}}=2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+4\hat{\mathrm{k}},\overrightarrow{\mathrm{q}}=3\hat{\mathrm{i}}+4\hat{\mathrm{j}}+5\hat{\mathrm{k}}\\ & \Rightarrow \overrightarrow{\mathrm{p}}\times \overrightarrow{\mathrm{q}}=\left|\begin{array}{lll}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|=-\hat{\mathrm{i}}+2\hat{\mathrm{j}}-\hat{\mathrm{k}}\\ & \mathrm{A}\equiv (1,2,3)B\equiv (\lambda ,4,5)\\ & \text{ Shortest Distance }=\left|\frac{\overrightarrow{\mathrm{AB}}\cdot (\overrightarrow{\mathrm{P}}\times \overrightarrow{\mathrm{q}})}{|\overrightarrow{\mathrm{P}}\times \overrightarrow{\mathrm{q}}|}\right|\end{array}$
$\begin{array}{rl}& \frac{1}{\sqrt{6}}=\left|\frac{((\lambda -1)\hat{i}+2\hat{j}+2\hat{k})\cdot (-\hat{i}+2\hat{j}-\hat{k})}{\sqrt{6}}\right|\\ & \Rightarrow |-\lambda +1+4-2|=1\Rightarrow |\lambda -3|=1\\ & \Rightarrow \lambda =3\pm 1=4,2\end{array}$
Radius of circle passing through points
$\begin{array}{rl}& (0,0),(4,2)\&(2,4)\\ & =\frac{\text{ abc }}{4\Delta }=\frac{\sqrt{20}\times \sqrt{20}\times \sqrt{8}}{4\times \frac{1}{2}\left|\begin{array}{lll}1& 1& 1\\ 0& 4& 2\\ 0& 2& 4\end{array}\right|}=\frac{20\times 2\sqrt{2}}{2\times 12}\\ & =\frac{5\sqrt{2}}{3}\end{array}$