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As far as I've understood $${dx \over P} = {dy \over Q} ={dz \over Q} \hspace{2 cm} (1) $$ Gives system of curves, $v= 0$ and $u=0$ be it's two solution. The solution of $(1)$ is the intersection of two surfaces $u=0 , \text{and } v=0 $.

But while solving $Pdx + Q dy +R dz = 0 \hspace{1cm} (2)$ using the method the auxiliary equation, we change it into form as $(1)$ $${dx \over {\partial Q \over \partial z} - {\partial R \over \partial y}}={dy \over {\partial R \over \partial x} - {\partial P \over \partial z}}={dz \over {\partial P \over \partial y} - {\partial Q \over \partial x}} \hspace{2cm} (3)$$

How does solution of $(3)$ relate to $(2)$ which is a curve while solution of $(3)$ is a surface?

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