Show that each of the given three vectors is a unit vector:
$\frac{1}{7}(2 \hat{i}+3 \hat{j}+6 \hat{k}), \frac{1}{7}(3 \hat{i}-6 \hat{j}+2 \hat{k}), \frac{1}{7}(6 \hat{i}+2 \hat{j}-3 \hat{k})$
Also,show that they are mutually perpendicular to each other.

Let $\vec{a}=\frac{1}{7}(2 \hat{i}+3 \hat{j}+6 \hat{k})=\frac{2}{7} \hat{i}+\frac{3}{7} \hat{j}+\frac{6}{7} \hat{k}$
$\vec{b}=\frac{1}{7}(3 \hat{i}-6 \hat{j}+2 \hat{k})=\frac{3}{7} \hat{i}-\frac{6}{7} \hat{j}+\frac{2}{7} \hat{k}$
$\vec{c}=\frac{1}{7}(6 \hat{i}+2 \hat{j}-3 \hat{k})=\frac{6}{7} \hat{i}+\frac{2}{7} \hat{j}-\frac{3}{7} \hat{k}$
$|\vec{a}|=\sqrt{(\frac{2}{7})^{2}+(\frac{3}{7})^{2}+(\frac{6}{7})^{2}}=\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}=\sqrt{\frac{49}{49}}=1$
$|\vec{b}|=\sqrt{(\frac{3}{7})^{2}+(-\frac{6}{7})^{2}+(\frac{2}{7})^{2}}=\sqrt{\frac{9}{49}+\frac{36}{49}+\frac{4}{49}}=\sqrt{\frac{49}{49}}=1$
$|\vec{c}|=\sqrt{(\frac{6}{7})^{2}+(\frac{2}{7})^{2}+(-\frac{3}{7})^{2}}=\sqrt{\frac{36}{49}+\frac{4}{49}+\frac{9}{49}}=\sqrt{\frac{49}{49}}=1$
Thus,each of the given three vectors is a unit vector.
$\vec{a} \cdot \vec{b}=\frac{2}{7} \times \frac{3}{7}+\frac{3}{7} \times(-\frac{6}{7})+\frac{6}{7} \times \frac{2}{7}=\frac{6}{49}-\frac{18}{49}+\frac{12}{49}=0$
$\vec{b} \cdot \vec{c}=\frac{3}{7} \times \frac{6}{7}+(-\frac{6}{7}) \times \frac{2}{7}+\frac{2}{7} \times(-\frac{3}{7})=\frac{18}{49}-\frac{12}{49}-\frac{6}{49}=0$
$\vec{c} \cdot \vec{a}=\frac{6}{7} \times \frac{2}{7}+\frac{2}{7} \times \frac{3}{7}+(-\frac{3}{7}) \times \frac{6}{7}=\frac{12}{49}+\frac{6}{49}-\frac{18}{49}=0$
Since the dot product of each pair is $0$,the given three vectors are mutually perpendicular to each other.