The value of $\lambda$ and $\mu$ if $(2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\vec{0}$ are respectively:

$3, \frac{27}{2}$

The correct answer is Option (1) - $3, \frac{27}{2}$

$\begin{vmatrix}\hat i&\hat j&\hat k\\2&6&27\\1&λ&μ\end{vmatrix}$

$⇒(6μ-27λ)\hat i+(-2μ+27λ)\hat j+(2λ-6)\hat k=\vec 0$

so $2λ=6$, $2μ=27$

$λ=3$, $μ=\frac{27}{2}$