Correct option is (2) \(4 \sqrt{3}\)
\(\frac{x-3}{2}=\frac{y+15}{-7}=\frac{z-9}{5} \,\&\, \frac{x+1}{2}=\frac{y-1}{1}=\frac{z-9}{-3}\)
S. D = \( =\frac{\left|\left(\overline{\mathrm{a}}_{2} \cdot \overline{\mathrm{a}}_{1}\right) \cdot\left(\overline{\mathrm{b}}_{1} \cdot \overline{\mathrm{b}}_{2}\right)\right|}{\left|\overline{\mathrm{b}}_{1} \times \overline{\mathrm{b}}_{2}\right|}\)
\(\mathrm{a}_{1}=3,-15,9\)
\( \mathrm{a}_{2}=-1,1,9\)
\( b_{1}=2,-7,5\)
\(\mathrm{b}_{2}=2,1,-3\)
\(a_{2}-a_{1}=-4,16,0\)
\(\overline{\mathrm{b}}_{1} \times \overline{\mathrm{b}}_{2}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & -7 & 5 \\ 2 & 1 & -3\end{array}\right|=\hat{\mathrm{i}}(16)-\hat{\mathrm{j}}(-16)+\hat{\mathrm{k}}(16)\)
\(16(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\)
\(\left|\overline{\mathrm{b}}_{1} \times \overline{\mathrm{b}}_{2}\right|=16 \sqrt{3}\)
\(\therefore\left(\overline{\mathrm{a}}_{2}-\overline{\mathrm{a}}_{1}\right) \cdot\left(\overline{\mathrm{b}}_{1}-\overline{\mathrm{b}}_{2}\right)=16[-4+16]=(16)(12)\)
\( S.D. =\frac{(16)(12)}{16 \sqrt{3}}=4 \sqrt{3}\)
