Since ${ }^{\mathbf{a}, \mathbf{b}}$ and $\mathbf{c}$ are mutually perpendicular, so $\mathbf{a} \cdot \mathbf{b}=\mathbf{b} \cdot \mathbf{c}=\mathbf{c} \cdot \mathbf{a}=\mathbf{0}$
Angle between $\mathbf{a}$ and $\mathbf{a}+\mathbf{b}+\mathbf{c}$ is
$\cos \theta=\frac{\mathbf{a} \cdot(\mathbf{a}+\mathbf{b}+\mathbf{c})}{|\mathbf{a} \| \mathbf{a}+\mathbf{b}+\mathbf{c}|}$
Now $|\mathbf{a}|=|\mathbf{b}|=|\mathbf{c}|=\mathbf{a}$
$\begin{aligned}
& |\mathbf{a}+\mathbf{b}+\mathbf{c}|^2=\mathbf{a}^2+\mathbf{b}^2+\mathbf{c}^2+2 \mathbf{a} \cdot \mathbf{b}+2 \mathbf{b} \cdot \mathbf{c}+2 \mathbf{c} . \\
& \quad=a^2+a^2+a^2+0+0+0 \\
& \Rightarrow|\mathbf{a}+\mathbf{b}+\mathbf{c}|^2=3 a^2 \Rightarrow \mathbf{a}+\mathbf{b}+\mathbf{c} \mid=\sqrt{3} a
\end{aligned}$
Putting this value in (i), we get $\quad \theta=\cos ^{-1} \frac{1}{\sqrt{3}}$.