Let S be the set of all real values of lambda | vedclass

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(B) Let the four points be $A(-\lambda^2, 1, 1)$,$B(1, -\lambda^2, 1)$,$C(1, 1, -\lambda^2)$,and $D(-1, -1, 1)$.Since the plane passes through all fou

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$\{\sqrt{3}, -\sqrt{3}\}$

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(B) Let the four points be $A(-\lambda^2, 1, 1)$,$B(1, -\lambda^2, 1)$,$C(1, 1, -\lambda^2)$,and $D(-1, -1, 1)$.Since the plane passes through all four points,they must be coplanar.The condition for four points $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,$(x_3, y_3, z_3)$,and $(x_4, y_4, z_4)$ to be coplanar is given by the determinant:$\begin{vmatrix} x_1-x_4 & y_1-y_4 & z_1-z_4 \ x_2-x_4 & y_2-y_4 & z_2-z_4 \ x_3-x_4 & y_3-y_4 & z_3-z_4 \end{vmatrix} = 0$Substituting the coordinates:$\begin{vmatrix} -\lambda^2+1 & 1+1 & 1-1 \ 1+1 & -\lambda^2+1 & 1-1 \ 1+1 & 1+1 & -\lambda^2-1 \end{vmatrix} = 0$$\begin{vmatrix} 1-\lambda^2 & 2 & 0 \ 2 & 1-\lambda^2 & 0 \ 2 & 2 & -(\lambda^2+1) \end{vmatrix} = 0$Expanding along the third column:$-(\lambda^2+1) \cdot [(1-\lambda^2)^2 - 4] = 0$$-(\lambda^2+1) \cdot (1-\lambda^2-2)(1-\lambda^2+2) = 0$$-(\lambda^2+1) \cdot (-1-\lambda^2)(3-\lambda^2) = 0$$(\lambda^2+1)^2 (3-\lambda^2) = 0$Since $\lambda$ is real,$\lambda^2+1 
eq 0$. Therefore,$3-\lambda^2 = 0$,which gives $\lambda^2 = 3$.Thus,$\lambda = \pm \sqrt{3}$.

Let $S$ be the set of all real values of $\lambda$ such that a plane passing through the points $(-\lambda^2, 1, 1)$,$(1, -\lambda^2, 1)$,and $(1, 1, -\lambda^2)$ also passes through the point $(-1, -1, 1)$. Then $S$ is equal to

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