Show that the lines : \[\frac{1-x}{2}=\frac{y-3}{4}=\frac{z}{-1} \text { and } \frac{x-4}{3}=\frac{2 y-2}{-4}=z-1 \text { are coplanar. } \]

Equations of given lines in standard form are \[\begin{aligned} & \frac{x-1}{-2}=\frac{y-3}{4}=\frac{z}{-1} \text { and } \\ & \frac{x-4}{3}=\frac{y-1}{-2}=\frac{z-1}{1} \end{aligned} \] Lines are coplanar if \[\left|\begin{array}{ccc} x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}\right|=0 \] \(\displaystyle \mathrm{LHS}=\left|\begin{array}{rrr} 3 & -2 & 1 \\ -2 & 4 & -1 \\ 3 & -2 & 1 \end{array}\right|\) \[\begin{aligned} & =3(4-2)+2(-2+3)+1(4-12) \\ & =6+2-8 \\ & =0 \end{aligned} \] ⇒ Coplanar.