Explanation:
B Given,
\(l_1: \frac{\mathrm{x}-3}{1}=\frac{\mathrm{y}-5}{-2}=\frac{\mathrm{z}-7}{1}\)
And, \(l_2: \frac{\mathrm{x}+1}{7}=\frac{\mathrm{y}+1}{-6}=\frac{\mathrm{z}+1}{1}\)
\(\left(\mathrm{x}_1, \mathrm{y}_1, \mathrm{z}_1=3,5,7\right)\)
\(\left(\mathrm{a}_1, \mathrm{~b}_1, \mathrm{c}_1=1,-2,1\right)\)
And,
\(\left(\mathrm{x}_2, \mathrm{y}_2, \mathrm{z}_2=-1,-1,-1\right)\)
\(\left(\mathrm{a}_2, \mathrm{~b}_2, \mathrm{c}_2=7,-6,1\right)\)
Shortest distance between two lines is -
\(\mathrm{d}=\frac{\left|\begin{array}{ccc}\mathrm{x}_2-\mathrm{x}_1 & \mathrm{y}_2-\mathrm{y}_1 & \mathrm{z}_2-\mathrm{z}_1 \\ \mathrm{a}_1 & \mathrm{~b}_1 & \mathrm{c}_1 \\ \mathrm{a}_2 & \mathrm{~b}_2 & \mathrm{c}_2\end{array}\right|}{\sqrt{\left(\mathrm{b}_1 \mathrm{c}_2-\mathrm{b}_2 \mathrm{c}_1\right)^2+\left(\mathrm{c}_1 \mathrm{a}_2-\mathrm{c}_2 \mathrm{a}_1\right)^2+\left(\mathrm{a}_1 \mathrm{~b}_2-\mathrm{a}_2 \mathrm{~b}_1\right)^2}}\)
\(\Rightarrow \mathrm{d}=\frac{\left|\begin{array}{ccc}-1-3 & -1-5 & -1-7 \\ 1 & -2 & 1 \\ 7 & -6 & 1\end{array}\right|}{\sqrt{(-2+6)^2+(7-1)^2+(-6+14)^2}}\)
\(\Rightarrow d =\left|\frac{-4(-2+6)+6(1-7)-8(-6+14)}{\sqrt{16+36+64}}\right|\)
\(\Rightarrow d =\left|\frac{-16-36-64}{\sqrt{116}}\right|=\left|\frac{-116}{\sqrt{116}}\right|=\frac{116}{\sqrt{116}}=\sqrt{116}\)
Hence,
\(\mathrm{d}=2 \sqrt{29} \text { units. }\)