$$\begin{aligned} \text{Let}\quad & \overrightarrow{\mathbf{a}}=l_1 \hat{\mathbf{i}}+m_1 \hat{\mathbf{j}}+n_1 \hat{\mathbf{k}} \\ & \overrightarrow{\mathbf{b}}=l_2 \hat{\mathbf{i}}+m_2 \hat{\mathbf{j}}+n_2 \hat{\mathbf{k}} \\ & \overrightarrow{\mathbf{c}}=l_3 \hat{\mathbf{i}}+m_3 \hat{\mathbf{j}}+n_3 \hat{\mathbf{k}} \\ & \overrightarrow{\mathbf{d}}=\left(l_1+l_2+l_3\right) \hat{\mathbf{i}}+\left(m_1+m_2+m_2\right) \hat{\mathbf{j}}+\left(n_1+n_2+n_3\right) \hat{\mathbf{k}} \end{aligned}$$

Also, let $\alpha, \beta$ and $\gamma$ are the angles between $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{d}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{d}}, \overrightarrow{\mathbf{c}}$ and $\overrightarrow{\mathbf{d}}$.

$$\begin{aligned} \therefore \quad \cos \alpha & =l_1\left(l_1+l_2+l_3\right)+m_1\left(m_1+m_2+m_3\right)+n_1\left(n_1+n_2+n_3\right) \\ & =l_1^2+l_1 l_2+l_1 l_3+m_1^2+m_1 m_2+m_1 m_3+n_1^2+n_1 n_2+n_1 n_3 \end{aligned}$$

$$\begin{aligned} & =\left(l_1^2+m_1^2+n_1^2\right)+\left(l_1 l_2+l_1 l_3+m_1 m_2+m_1 m_3+n_1 n_2+n_1 n_3\right) \\ & =1+0=1 \\ & {\left[\because l_1^2+m_1^2+n_1^2=1 \text { and } l_1 \perp l_2, l_1 \perp l_3, m_1 \perp m_2, m_1 \perp m_3, n_1 \perp n_2, n_1 \perp n_3\right]} \end{aligned}$$

Similarly,

$$\begin{aligned} \cos \beta & =l_2\left(l_1+l_2+l_3\right)+m_2\left(m_1+m_2+m_3\right)+n_2\left(n_1+n_2+n_3\right) \\ & =1+0 \text { and } \cos \gamma=1+0 \end{aligned}$$

$$\Rightarrow \quad \cos \alpha=\cos \beta=\cos \gamma$$

$$\Rightarrow \quad \alpha=\beta=\gamma$$

So, 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 the three mutually perpendicular lines whose direction cosines are $l_1, m_1, n_1, l_2, m_2, n_2$ and $l_3, m_3, n_3$ respectively.