$(2 \hat{i}+6 \hat{i}+27 \hat{k}) imes(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overline{0}$, then $\lambda$ and $\mu$ are respectively

Question

$(2 \hat{i}+6 \hat{i}+27 \hat{k}) 	imes(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overline{0}$, then $\lambda$ and $\mu$ are respectively

$(2 \hat{i}+6 \hat{i}+27 \hat{k}) \times(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overline{0}$, then $\lambda$…