Correct Answer - Option 2 :
\(r \ge 5\sqrt {\frac{3}{2}}\)
Given, \(\vec a = 3\hat i + 2\hat j + x\hat k{\rm{\;and\;}}\vec b = \hat i - \hat j + \hat k\)
Now, \(\vec a \times \vec b = \left| {\begin{array}{*{20}{c}}
{\hat i}&{\hat j}&{\hat k}\\
3&2&x\\
1&{ - 1}&1
\end{array}} \right|\)
\(= \left( {2 + x} \right)\hat i + \left( {x - 3} \right)\hat j - 5\hat k\)
\(\left| {\vec a \times \vec b} \right| = \sqrt {{{(2 + x)}^2} + {{(x - 3)}^2} + {{( - 5)}^2}} = r\)
\(\Rightarrow r = \sqrt {4 + {x^2} + 4x + {x^2} + 9 - 6x + 25}\)
\(= \sqrt {2{x^2} - 2x + 38}\)
Add \(\frac{1}{2}{\rm{\;and\;}} - \frac{1}{2}\) in above equation, we get
\(\Rightarrow r = \sqrt {2{x^2} - 2x + \frac{1}{2} + 38 - \frac{1}{2}}\)
\(= \sqrt {2\left( {{x^2} - x + \frac{1}{4}} \right) + 38 - \frac{1}{2}}\)
\(= \sqrt {2{{\left( {x - \frac{1}{2}} \right)}^2} + \frac{{75}}{2}}\)
\(\Rightarrow r \ge \sqrt {\frac{{75}}{2}}\)
\(\Rightarrow r \ge \sqrt {\frac{{25 \times 3}}{2}}\)
Hence,
\(r \ge 5\sqrt {\frac{3}{2}}\)