The three straight lines $\frac{px}{l} = \frac{qy}{m} = \frac{rz}{n}$ , $\frac{x}{l} = \frac{y}{m} = \frac{z}{n}$ and $\frac{x}{pl} = \frac{y}{qm} = \frac{z}{rn}$ I need to find out the condition of their coplanarity. I'm seeing the solution as $p = q$ or $q = r$ or $r = s$
I can't find a way of getting a solution. Any kind of hint or guide would be helpful.
All three of your lines pass through the origin $(0,0,0)$, so you might as well describe them by their direction vectors, which I'd write without divisions as
$$\begin{pmatrix}lqr\\mpr\\npq\end{pmatrix}\qquad \begin{pmatrix}l\\m\\n\end{pmatrix}\qquad \begin{pmatrix}pl\\qm\\rn\end{pmatrix}$$
Now these three direction vectors lie in a single plane if the space spanned by them has dimension less than three, i.e. if the matrix formed by these has rank less than three, i.e. if the determinant of that matrix is zero. This is equivalent to the triple product amd mentioned in a comment.
$$0\overset!=\begin{vmatrix}lqr&l&pl\\mpr&m&qm\\nqp&n&rn\end{vmatrix} =-l\,m\,n\,(p-r)(q-r)(p-q)$$
So if all your parameters are non-zero (to avoid division by zero in the last of your equations), then $p=q$ and $q=r$ are two solutions which you already observed. $p=r$ is the third one. The $r=s$ you mention makes little sense as there is no $s$ in the formulas you gave.