Q.If the lines $\dfrac{x-x_1}{l_1}=\dfrac{y-y_1}{m_1}=\dfrac{z-z_1}{n_1}$ and $\dfrac{x-x_2}{l_2}=\dfrac{y-y_2}{m_2}=\dfrac{z-z_2}{n_2}$ lie on the same plane, then write the number of ways to find the Cartesian equation of the above plane and explain in detail.

Explains that a plane containing two coplanar lines can be written using either line's point together with both direction ratios, giving two equivalent Cartesian forms.

1. The two lines $\dfrac{x-x_1}{l_1}=\dfrac{y-y_1}{m_1}=\dfrac{z-z_1}{n_1}$ and $\dfrac{x-x_2}{l_2}=\dfrac{y-y_2}{m_2}=\dfrac{z-z_2}{n_2}$ are given to be coplanar, so a single plane contains both of them.
2. A plane containing both lines must contain both direction vectors $(l_1,m_1,n_1)$ and $(l_2,m_2,n_2)$, so its normal is perpendicular to both, i.e. $\vec n=(l_1,m_1,n_1)\times(l_2,m_2,n_2)$.
3. Method 1 — anchor at $(x_1,y_1,z_1)$: since $(x_1,y_1,z_1)$ lies on line 1, and hence on the plane, the plane's equation is the point-normal (scalar triple product) form $\begin{vmatrix}x-x_1&y-y_1&z-z_1\\l_1&m_1&n_1\\l_2&m_2&n_2\end{vmatrix}=0$.
4. Method 2 — anchor at $(x_2,y_2,z_2)$: equally, since $(x_2,y_2,z_2)$ lies on line 2, and hence on the same plane, the equation can instead be written as $\begin{vmatrix}x-x_2&y-y_2&z-z_2\\l_1&m_1&n_1\\l_2&m_2&n_2\end{vmatrix}=0$. …