If the vectors \(\bar{a}=\hat{i}-\hat{j}+2 \hat{k}, \bar{b}=2 \hat{i}+4 \hat{j}+\hat{k}\) and \(\overline{\mathrm{c}}=\mathrm{mi}+\mathrm{j}+\mathrm{nk}\) are mutually perpendicular, then \((\mathrm{m}, \mathrm{n})\) is

\(\overline{\mathrm{a}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}, \overline{\mathrm{~b}}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\hat{\mathrm{k}} \text { and } \overline{\mathrm{c}}=m \hat{\mathrm{i}}+\hat{\mathrm{j}}+\mathrm{n} \hat{\mathrm{k}}\)
\(\overline{\mathrm{a}}\) and \(\overline{\mathrm{c}}\) are orthogonal
\(\begin{aligned}
\therefore \quad & \overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=0 \\
& \mathrm{~m}(1)+(1)(-1)+\mathrm{n}(2)=0 \\
& m-1+2 \mathrm{n}=0 \\
& m+2 n=1...(i)
\end{aligned}\)
Also, \(\overline{\mathrm{b}}\) and \(\overline{\mathrm{c}}\) are orthogonal
\(\begin{array}{ll}
\therefore \quad & \bar{b} \cdot \overline{\mathrm{c}}=0 \\
& 2(\mathrm{~m})+4(1)+\mathrm{n}(1)=0 \\
& 2 \mathrm{~m}+\mathrm{n}=-4...(ii)
\end{array}\)
Solving (i) and (ii), we get
\(\begin{array}{ll}
& m=-3, \mathrm{n}=2 \\
\therefore \quad & (\mathrm{~m}, \mathrm{n})=(-3,2)
\end{array}\)