Show that the vector $\hat{i} + \hat{j} + \hat{k}$ is equally inclined with the coordinate axes.
Let $\overrightarrow{p} = \hat{i} + \hat{j} + \hat{k}$
Let $\overrightarrow{p}$ make an angle $\alpha$ with the X axis, $\beta$ with the Y axis and $\gamma$ with the Z axis.
Unit vector along the X axis is $= \hat{x} = 1 \hat{i} + 0 \hat{j} + 0 \hat{k}$
Unit vector along the Y axis is $= \hat{y} = 0 \hat{i} + 1 \hat{j} + 0 \hat{k}$
Unit vector along the Z axis is $= \hat{z} = 0 \hat{i} + 0 \hat{j} + 1 \hat{k}$
Now, $\left|\overrightarrow{p}\right| = \sqrt{\left(1\right)^2 + \left(1\right)^2 + \left(1\right)^2} = \sqrt{3}$
Angle between $\overrightarrow{p}$ and the X axis is the angle between $\overrightarrow{p}$ and $\hat{x}$.
$\overrightarrow{p} \cdot \hat{x} = \left|\overrightarrow{p}\right| \left|\hat{x}\right| \cos \alpha$
$\left|\hat{x}\right| = 1$
$\begin{aligned}
\therefore \; \cos \alpha & = \dfrac{\overrightarrow{p} \cdot \hat{x}}{\left|\overrightarrow{p}\right| \left|\hat{x}\right|} \\\\
& = \dfrac{\left(\hat{i} + \hat{j} + \hat{k}\right) \cdot \left(\hat{i} + 0 \hat{j} + 0 \hat{k}\right)}{\sqrt{3} \times 1} \\\\
& = \dfrac{1}{\sqrt{3}} \;\;\; \cdots \; (1)
\end{aligned}$
Angle between $\overrightarrow{p}$ and the Y axis is the angle between $\overrightarrow{p}$ and $\hat{y}$.
$\overrightarrow{p} \cdot \hat{y} = \left|\overrightarrow{p}\right| \left|\hat{y}\right| \cos \beta$
$\left|\hat{y}\right| = 1$
$\begin{aligned}
\therefore \; \cos \beta & = \dfrac{\overrightarrow{p} \cdot \hat{y}}{\left|\overrightarrow{p}\right| \left|\hat{y}\right|} \\\\
& = \dfrac{\left(\hat{i} + \hat{j} + \hat{k}\right) \cdot \left(0 \hat{i} + \hat{j} + 0 \hat{k}\right)}{\sqrt{3} \times 1} \\\\
& = \dfrac{1}{\sqrt{3}} \;\;\; \cdots \; (2)
\end{aligned}$
Angle between $\overrightarrow{p}$ and the Z axis is the angle between $\overrightarrow{p}$ and $\hat{z}$.
$\overrightarrow{p} \cdot \hat{z} = \left|\overrightarrow{p}\right| \left|\hat{z}\right| \cos \gamma$
$\left|\hat{z}\right| = 1$
$\begin{aligned}
\therefore \; \cos \gamma & = \dfrac{\overrightarrow{p} \cdot \hat{z}}{\left|\overrightarrow{p}\right| \left|\hat{z}\right|} \\\\
& = \dfrac{\left(\hat{i} + \hat{j} + \hat{k}\right) \cdot \left(0 \hat{i} + 0 \hat{j} + \hat{k}\right)}{\sqrt{3} \times 1} \\\\
& = \dfrac{1}{\sqrt{3}} \;\;\; \cdots \; (3)
\end{aligned}$
$\therefore$ $\;$ From equations $(1)$, $(2)$ and $(3)$, $\cos \alpha = \cos \beta = \cos \gamma$
$\implies$ $\alpha = \beta = \gamma$
i.e. $\;$ $\overrightarrow{p}$ is equally inclined with the coordinate axes.