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

Let $\vec a = \cfrac{1}{7}(2\hat j + 3\hat j + 6\hat k),\vec b = \cfrac{1}{7}(3\hat i - 6\hat j + 2\hat k)$ and $\vec c = \cfrac{1}{7}(6\hat i + 2\hat j - 3\hat k)$

$\therefore$ $|\vec a| = \sqrt {{{\left( {\cfrac{2}{7}} \right)}^2} + {{\left( {\cfrac{3}{7}} \right)}^2} + {{\left( {\cfrac{6}{7}} \right)}^2}}$
$= \sqrt {\cfrac{4}{{49}} + \cfrac{9}{{49}} + \cfrac{{36}}{{46}}} = \sqrt {\cfrac{{49}}{{49}}} = \sqrt 1 = 1$

$|\vec b| = \sqrt {{{\left( {\cfrac{3}{7}} \right)}^2} + {{\left( {\cfrac{{ - 6}}{7}} \right)}^2} + {{\left( {\cfrac{2}{7}} \right)}^2}} = \sqrt {\cfrac{9}{{49}} + \cfrac{{36}}{{49}} + \cfrac{4}{{49}}} = \sqrt {\cfrac{{49}}{{49}}} = \sqrt 1 = 1$

$|\vec c| = \sqrt {{{\left( {\cfrac{6}{7}} \right)}^2} + {{\left( {\cfrac{2}{7}} \right)}^2} + {{\left( {\cfrac{{ - 3}}{7}} \right)}^2}} = \sqrt {\cfrac{{49}}{{49}}} = \sqrt 1 = 1$

Hence $\vec a,\vec b,\vec c$ are unit vectors.
Now, $\vec a \cdot \vec b = \cfrac{1}{{49}}[(2)(3) + (3)( - 6) + (6)(2)] = \cfrac{1}{{49}}[6 - 18 + 12] = 0$

So, $\vec a$ is perpendicular to $\vec b$.
$\vec b \cdot \vec c = \cfrac{1}{{49}}[(3).(6) + ( - 6)(2) + (2)( - 3)] = \cfrac{1}{{49}}[18 - 12 - 6] = 0$
So, $\vec b$ is perpendicular to $\vec c$.

$\vec c.\vec a = \cfrac{1}{{49}}[(6)(2) + (2)(3) + ( - 3)(6)] = \cfrac{1}{{49}}[12 + 6 - 18] = 0$
So, $\vec c$ is perpendicular to $\vec a$.

Hence, $\vec a,\vec b$ and $\vec c$ are three mutually perpendicular unit vectors.