The unit vector perpendicular to each of the vectors \(\bar{a}+\bar{b}\) and \(\bar{a}-\bar{b}\), where \(\bar{a}=\hat{i}+\hat{j}+\hat{k}\) and \(\overline{\mathrm{b}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}\) is

\(\begin{aligned} \bar{a}+\bar{b} & =(\hat{i}+\hat{j}+\hat{k})+(3 \hat{i}-2 \hat{j}+5 \hat{k}) \\ & =4 \hat{i}-\hat{j}+6 \hat{k} \\ \bar{a}-\bar{b} & =(\hat{i}+\hat{j}+\hat{k})-(3 \hat{i}-2 \hat{j}+5 \hat{k}) \\ & =-2 \hat{i}+3 \hat{j}-4 \hat{k}\end{aligned}\)
\(\therefore \quad\) Vector perpendicular to \((\bar{a}+\bar{b})\) and \((\bar{a}-\bar{b})\) is
\(\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
4 & -1 & 6 \\
-2 & 3 & -4
\end{array}\right|=-14 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}\)
\(\therefore \quad\) Required unit vector is
\(\frac{-14 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}}{\sqrt{(-14)^2+4^2+(10)^2}}=\frac{-14 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}}{\sqrt{312}}\)