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