The values of \(\lambda\) and \(\mu\) for which \(\left(2 \hat{i}+6\hat{j}+27\hat{k}\right)\times \left(\hat{i}+\lambda \hat{j}+\mu \hat{k}\right)=\vec{0}\) are

\(\lambda=3,\mu=\frac{27}{2}\) \(\lambda=\frac{27}{2},\mu=3\) \(\lambda=3,\mu=3\) \(\lambda=\frac{27}{2},\mu=\frac{27}{2}\)

\(\lambda=3,\mu=\frac{27}{2}\)

\(\begin{aligned}\text{Let }\vec{a}=2\hat{i}+6\hat{j}+27\hat{k},\vec{b}&=\hat{i}+\lambda \hat{j}+\mu \hat{k}\\ \vec{a}\times \vec{b}=\left|\begin{array}{lll}\hat{i}&\hat{j}&\hat{k}\\ 2&6&27\\ 1&\lambda & \mu \end{array}\right|&=0\\ (6\mu-27\lambda)\hat{i}-(2\mu -27)\hat{j}+(2\lambda -6)\hat{k}&=0\\ 6\mu -27 \lambda =0, 2\mu -27=0, 2\lambda -6&=0\\ \mu=\frac{27}{6}\lambda, \mu=\frac{27}{2},\lambda &=3\\ \lambda=3, \mu& =\frac{27}{2}\end{aligned}\)