The number of values of lambda for which the | vedclass

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(C) Three points $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ are collinear if the area of the triangle formed by them is $0$.$\frac{1}{2} \left| \begin

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$2$

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(C) Three points $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ are collinear if the area of the triangle formed by them is $0$.$\frac{1}{2} \left| \begin{matrix} \lambda+1 & 1 & 1 \ 2\lambda+1 & 3 & 1 \ 2\lambda+2 & 2\lambda & 1 \end{matrix} ight| = 0$Applying row operations $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$:$\left| \begin{matrix} \lambda+1 & 1 & 1 \ \lambda & 2 & 0 \ \lambda+1 & 2\lambda-1 & 0 \end{matrix} ight| = 0$Expanding along the third column:$1 \cdot [\lambda(2\lambda-1) - 2(\lambda+1)] = 0$$2\lambda^2 - \lambda - 2\lambda - 2 = 0$$2\lambda^2 - 3\lambda - 2 = 0$$2\lambda^2 - 4\lambda + \lambda - 2 = 0$$2\lambda(\lambda - 2) + 1(\lambda - 2) = 0$$(2\lambda + 1)(\lambda - 2) = 0$Thus,$\lambda = -1/2$ or $\lambda = 2$.There are $2$ possible values for $\lambda$.

The number of values of $\lambda$ for which the points $(\lambda + 1, 1)$,$(2\lambda + 1, 3)$,and $(2\lambda + 2, 2\lambda)$ are collinear is:

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