If $l_{1}, m_{1}, n_{1}; l_{2}, m_{2}, n_{2}; l_{3}, m_{3}, n_{3}$ are the direction cosines of three mutually perpendicular lines,prove that the line whose direction cosines are proportional to $l_{1}+l_{2}+l_{3}, m_{1}+m_{2}+m_{3}, n_{1}+n_{2}+n_{3}$ makes equal angles with them.

(N/A) Let $\vec{a} = l_{1}\hat{i} + m_{1}\hat{j} + n_{1}\hat{k}$,$\vec{b} = l_{2}\hat{i} + m_{2}\hat{j} + n_{2}\hat{k}$,and $\vec{c} = l_{3}\hat{i} + m_{3}\hat{j} + n_{3}\hat{k}$ be the unit vectors along the three mutually perpendicular lines.
Let $\vec{d} = (l_{1}+l_{2}+l_{3})\hat{i} + (m_{1}+m_{2}+m_{3})\hat{j} + (n_{1}+n_{2}+n_{3})\hat{k} = \vec{a} + \vec{b} + \vec{c}$.
Let $\alpha, \beta, \gamma$ be the angles between $\vec{d}$ and $\vec{a}, \vec{b}, \vec{c}$ respectively.
Then $\cos \alpha = \frac{\vec{a} \cdot \vec{d}}{|\vec{a}| |\vec{d}|} = \frac{\vec{a} \cdot (\vec{a} + \vec{b} + \vec{c})}{|\vec{d}|} = \frac{\vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}}{|\vec{d}|}$.
Since the lines are mutually perpendicular,$\vec{a} \cdot \vec{b} = 0$,$\vec{a} \cdot \vec{c} = 0$,and $\vec{a} \cdot \vec{a} = 1$.
Thus,$\cos \alpha = \frac{1}{|\vec{d}|}$.
Similarly,$\cos \beta = \frac{\vec{b} \cdot (\vec{a} + \vec{b} + \vec{c})}{|\vec{d}|} = \frac{1}{|\vec{d}|}$ and $\cos \gamma = \frac{\vec{c} \cdot (\vec{a} + \vec{b} + \vec{c})}{|\vec{d}|} = \frac{1}{|\vec{d}|}$.
Since $\cos \alpha = \cos \beta = \cos \gamma$,it follows that $\alpha = \beta = \gamma$. Hence,the line makes equal angles with the three mutually perpendicular lines.