We have $|\vec{a}|=|\vec{b}|=|\vec{c}|=\lambda$
Also $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ are mutually perpendicular
$$
\vec{a} \cdot \vec{b}=0, \vec{b} \cdot \vec{c}=0, \vec{c} \cdot \vec{a}=0
$$
let $\theta$ be the angle between vectors $\vec{a}$ and $\vec{a}+\vec{b}+\vec{c}$
$$
\begin{aligned}
&\therefore \quad \vec{a} \cdot(\vec{a}+\vec{b}+\vec{c})=|\vec{a}||\vec{a}+\vec{b}+\vec{c}| \cos \theta \\
&=\lambda|\vec{a}+\vec{b}+\vec{c}| \cos \theta
\end{aligned}
$$
Also $\overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}$
$$
\begin{aligned}
&=|\vec{a}|^2+0+0=\lambda^2 \\
&\text { from (i) },|\vec{a}|=\lambda^2, \text { from (ii) } \vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}=0 \\
&\Rightarrow \lambda^2=\lambda|\vec{a}+\vec{b}+\vec{c}| \cos \theta \\
&\Rightarrow \theta=\cos ^{-1} \frac{\lambda}{|\vec{a}+\vec{b}+\vec{c}|}
\end{aligned}
$$
Similarly agnel $\theta$ between $\vec{b}$ and $(\vec{a}+\vec{b}+\vec{c})$
$$
=\cos ^{-1} \frac{\lambda}{|\vec{a}+\vec{b}+\vec{c}|}
$$
and the angle between $\vec{c}$ and $(\vec{a}+\vec{b}+\vec{c})$
$$
=\cos ^{-1} \frac{\lambda}{|\vec{a}+\vec{b}+\vec{c}|}
$$
Thus $\vec{a}+\vec{b}+\vec{c}$ is equally inclined with $\vec{a}, \vec{b}$ and $\vec{c}$