If the points $(2,-3),(\lambda,-1)$ and $(0,4)$ are collinear, then the value of $\lambda$ is :

$\frac{10}{7}$

$A(2, -3), B(\lambda,-1), C(0,4)$ → Collinear

$\Rightarrow P(2,-3,1), Q(\lambda,-1,1), R(0,4,1)$ → also collinear

$\Rightarrow\left|\begin{array}{rrr}2 & -3 & 1 \\ \lambda & -1 & 1 \\ 0 & 4 & 1\end{array}\right|=0 \Rightarrow \begin{aligned} & R_2 \rightarrow R_2-R_3 \\ & R_1 \rightarrow R_1-R_3\end{aligned}\left|\begin{array}{ccc}2 & -7 & 0 \\ \lambda & -5 & 0 \\ 0 & 4 & 1\end{array}\right|=0$

so  $1\left|\begin{array}{cc}2 & -7 \\ \lambda & -5\end{array}\right|=0 \Rightarrow -10+ 7 \lambda = 0$

so  $\lambda = \frac{10}{7}$

Option: 4