If the lines $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$ and $\frac{x-x_2}{p}=\frac{y-y_2}{q}=\frac{z-z_2}{r}$ lie on the same plane, then:

Two lines are coplanar if the vector connecting a point on the first line to a point on the second line, and the two direction vectors of the lines, are all linearly dependent.Point on line 1: $(x_1,y_1,z_1)$ Point on line 2: $(x_2,y_2,z_2)$ Direction vectors: $\overrightarrow{d_1}=(a,b,c)$ and $\overrightarrow{d_2}=(p,q,r)$.The condition for them to be coplanar is that their scalar triple product is zero:
$$\left|\begin{array}{ccc}{x}_1-x_2& y_1-y_2& z_1-z_2\\ a& b& c\\ p& q& r\end{array}\right|=0$$