Find $\lambda$ and $\mu$ if $(2\hat i + 6\hat j + 27\hat k) \times (\hat i + \lambda \widehat j + \mu \hat k) = \vec 0$.

Let $\vec a = 2\hat i + 6\hat j + 27\widehat k$ and $\vec b = \hat i + \lambda \hat j + \mu \hat k$

$\therefore$ $\vec a \times \vec b = \left| {\begin{array}{llllllllllllllllllll}{\hat i}&{\hat j}&{\hat k}\\2&6&{27}\\l&\lambda &\mu \end{array}} \right|$

$= \hat i(6\mu - 27\lambda ) - \hat j(2\mu - 27) + \hat k(2\lambda - 6)$

$= (6\mu - 27\lambda )\hat i + (27 - 2\mu )\hat j + (2\lambda - 6)\hat k$

By the question, $\vec a \times \vec b = \vec 0$

$\Rightarrow (6\mu - 27\lambda )\hat i + (27 - 2\mu )\hat j + (2\lambda - 6)\hat k = \vec 0$

$\Rightarrow 6\mu - 27\lambda = 0,(2\lambda - 6) = 0,(27 - 2\mu ) = 0$

$\Rightarrow \lambda = 3$ and $\mu = \cfrac{{27}}{2}$