Let ABC be a triangle whose vertices, $A(1,-1),B(0,2)C\left(x^{\mathrm{\prime }},y^{\mathrm{\prime }}\right)$ such that area of $\mathrm{△}ABC\text{ is }5\mathrm{sq}.$ and $C\left(x^{\mathrm{\prime }},y^{\mathrm{\prime }}\right)\text{ lie on }3x+y-4\lambda =0.$ $\text{ Then, }\lambda =$

It is given that area of $\mathrm{△}ABC\text{ is }5$ sq. units

$\begin{array}{l}\therefore \frac{1}{2}\left|\begin{array}{ccc}1& -1& 1\\ 0& 2& 1\\ x^{\mathrm{\prime }}& y^{\mathrm{\prime }}& 1\end{array}\right|=\pm 5\\ \Rightarrow \left(2-y^{\mathrm{\prime }}\right)+\left(0-x^{\mathrm{\prime }}\right)+\left(0-2x^{\mathrm{\prime }}\right)=\pm 10\\ \Rightarrow -3x^{\mathrm{\prime }}-y^{\mathrm{\prime }}+2=\pm 10\\ \Rightarrow 3x^{\mathrm{\prime }}+y^{\mathrm{\prime }}-2=\pm 10\\ \Rightarrow 3x^{\mathrm{\prime }}+y^{\mathrm{\prime }}-12=0\text{ or }3x^{\mathrm{\prime }}+y^{\mathrm{\prime }}+8=0\end{array}$

Point $C\left(x^{\mathrm{\prime }},y^{\mathrm{\prime }}\right)\text{ lies on }3x+y-4\lambda =0.\text{ Therefore, }\lambda =3$