Show that the vector

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$

Then,

$|\vec{a}|=\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}$

Therefore, the direction cosines of $\vec{a}$ are $\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$

Now, let $\alpha, \beta$, and $\gamma$ be the angles formed by $\vec{a}$ with the positive directions of $x, y$, and $z$ axes.

Then, we have $\cos \alpha=\frac{1}{\sqrt{3}}, \cos \beta=\frac{1}{\sqrt{3}}, \cos \gamma=\frac{1}{\sqrt{3}}$.

Hence, the given vector is equally inclined to axes OX, OY, and OZ.