OABC is a tetrahedron in which O is the origin and position vector of points A, B, C are $\hat{i}+2 \hat{j}+3 \hat{k}, 2 \hat{i} +\alpha \hat{j} + \hat{k} and \hat{i} + 3 \hat{j} +2 \hat{k}$ respectively. A value of ‘$\mathrm{\alpha }$ ’ for which shortest distance between OA and BC is $\sqrt{\frac{3}{2}}$

Unit vector $\left(\hat{\mathrm{n}}\right)$ perpendicular to $$\begin{gathered} \overrightarrow{\mathrm{OA}}\mathrm{and} \overrightarrow{\mathrm{CB}}\mathrm{is} \mathrm{in} \hat{\mathrm{n}} =\frac{\left(3\mathrm{\alpha }-7\right) \hat{i}-4 \hat{j}+\left(5-\mathrm{\alpha }\right) \hat{k}}{\sqrt{{\left(3\mathrm{\alpha }-7\right)}^2+16+{\left(5-\mathrm{\alpha }\right)}^2}} \\ \therefore \sqrt{\frac{3}{2}}=\left|\overline{\mathrm{BA}} \hat{\mathrm{n}}\right|\Rightarrow \mathrm{\alpha }=\frac{9}{7}\mathrm{or} 3 \end{gathered}$$

$\begin{array}{l}\overrightarrow{OA}=\hat{i}+2 \hat{j}+3 \hat{k},\overrightarrow{OB}=2 \hat{i}+\mathrm{\alpha } \hat{\mathrm{j}}+\hat{k}\text{ and }\overrightarrow{OC}=\hat{i}+3 \hat{j}+2 \hat{k}\\ \Rightarrow \overrightarrow{BA}=\overrightarrow{OA}-\overrightarrow{OB}=-\hat{i}+(2-\alpha ) \hat{j}+2 \hat{k},\\ \overrightarrow{CB}=\overrightarrow{OB}-\overrightarrow{OC}=\hat{i}+(\alpha -3) \hat{j}-\hat{k}\\ , \mathrm{and} \\ \text{ Unit vector ( }\widehat{n}\text{ ) perpendicular to }\overrightarrow{OA}\text{ and }\overrightarrow{CB}\text{ is }\widehat{n}=\frac{\overrightarrow{\mathrm{CB}}\times \overrightarrow{\mathrm{OA}}}{|\overrightarrow{\mathrm{CB}}\times \overrightarrow{\mathrm{OA}}|}\\ =\frac{\left(3\mathrm{\alpha }-7\right) \hat{i}-4 \hat{j}+\left(5-\mathrm{\alpha }\right) \hat{k}}{\sqrt{{\left(3\mathrm{\alpha }-7\right)}^2+16+{\left(5-\mathrm{\alpha }\right)}^2}}\\ \text{ As, }\overrightarrow{CB}\times \overrightarrow{OA}=\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ 1& \alpha -3& -1\\ 1& 2& 3\end{array}\right|=(3\alpha -7)\hat{i}-4\hat{j}+(5-\alpha )\hat{k}\\ \begin{array}{l}Shortest dis\tan ce=d=|\overrightarrow{BA}\cdot \widehat{n}|\\ \Rightarrow \sqrt{\frac{3}{2}}=\left|\frac{(7-3a)-4(2-\alpha )+2(5-\alpha )}{\sqrt{(3\alpha -7{)}^2+(16)+(5-\alpha {)}^2}}\right|\\ \Rightarrow \sqrt{\frac{3}{2}}=\frac{|-\alpha +9|}{\sqrt{10{\alpha }^2-52\alpha +90}}\\ squaring and simplifying\\ \begin{array}{r}7{\alpha }^2-30\alpha +27=0\\ \Rightarrow (7\alpha -9)(\alpha -3)=0\\ \Rightarrow \alpha =\frac{9}{7},3\end{array}\end{array}\\ \end{array}$