$$\begin{gathered} Let the lines \frac{x-1}{\lambda }=\frac{y-2}{1}=\frac{z-3}{2} and \\ \frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda } be coplanar and P be \\ the plane containing these two lines. Then which \\ of the following points does NOT lies on P? \end{gathered}$$

$$\begin{gathered} Given, L_1:\frac{x-1}{\lambda }=\frac{y-2}{1}=\frac{z-3}{2} \\ and L_2:\frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda } \\ are coplanar \\ \Rightarrow \left|\begin{array}{ccc}27& 20& 31\\ \lambda & 1& 2\\ -2& 3& \lambda \end{array}\right|=0\Rightarrow {\lambda }^2+9-6\lambda =0 \\ Normal to the plane P which contains given lines is given by \\ \Rightarrow {\left(\lambda -3\right)}^2=0 \\ \Rightarrow \lambda =3 \\ Now, normal of plane P, which contains L_1 and L_2 \\ \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 1& 2\\ -2& 3& 3\end{array}\right|=-3\hat{i}-13\hat{j}+11\hat{k} \\ \Rightarrow Equation of required plane P is-3\left(x-1\right)-13\left(y-2\right) +11\left(z-3\right)=0 \\ \Rightarrow 3x+13y-11z+4=0 \\ \therefore \left(0, 4, 5\right) does not lie on plane P. \end{gathered}$$