We have, $$2 l+2 m-n=0\quad\text{.... (i)}$$

$$\begin{aligned} & \text { and } \quad m n+n l+l m=0 \quad\text{.... (ii)}\\ & \text { Eliminating } m \text { from the both equations, we get } \end{aligned}$$

$$m=\frac{n-2 l}{2}\quad\text{[from Eq. (i)]}$$

$$\begin{array}{lr} \Rightarrow & \left(\frac{n-2 l}{2}\right) n+n l+l\left(\frac{n-2 l}{2}\right)=0 \\ \Rightarrow & \frac{n^2-2 n l+2 n l+n l-2 l^2}{2}=0 \\ \Rightarrow & n^2+n l-2 l^2=0 \\ \Rightarrow & n^2+2 n l-n l-2 l^2=0 \\ \Rightarrow & (n+2 l)(n-l)=0 \end{array}$$

$$\begin{array}{ll} \Rightarrow & n=-2 l \text { and } n=l \\ \therefore & m=\frac{-2 l-2 l}{2}, m=\frac{l-2 l}{2} \\ \Rightarrow & m=-2 l, m=\frac{-l}{2} \end{array}$$

Thus, the direction ratios of two lines are proportional to $l,-2 l,-2$ and $l, \frac{-l}{2}, l$.

$$\begin{array}{ll} \Rightarrow & 1,-2,-2 \text { and } 1, \frac{-1}{2}, 1 \\ \Rightarrow & 1,-2,-2 \text { and } 2,-1,2 \end{array}$$

Also, the vectors parallel to these lines are $\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$, respectively.

$$\begin{array}{ll} \therefore \quad \cos \theta & =\frac{\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}|}=\frac{(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \cdot(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})}{3 \cdot 3} \\ & =\frac{2+2-4}{9}=0 \\ \therefore \quad \theta & =\frac{\pi}{2} \quad\left[\because \cos \frac{\pi}{2}=0\right] \end{array}$$

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

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.