If S is the set of all real values of a such | vedclass

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(B) Let the four points be $A(-a^2, 1, 1)$,$B(1, -a^2, 1)$,$C(1, 1, -a^2)$,and $D(-1, -1, 1)$.Since these four points are coplanar,the volume of the t

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

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(B) Let the four points be $A(-a^2, 1, 1)$,$B(1, -a^2, 1)$,$C(1, 1, -a^2)$,and $D(-1, -1, 1)$.Since these four points are coplanar,the volume of the tetrahedron formed by them is zero,or the determinant of the vectors formed by them is zero.Consider the vectors $\vec{AB} = (1+a^2, -a^2-1, 0)$,$\vec{AC} = (1+a^2, 0, -a^2-1)$,and $\vec{AD} = (-1+a^2, -2, 0)$.The condition for coplanarity is $\det(\vec{AB}, \vec{AC}, \vec{AD}) = 0$.$\begin{vmatrix} 1+a^2 & -(1+a^2) & 0 \ 1+a^2 & 0 & -(1+a^2) \ a^2-1 & -2 & 0 \end{vmatrix} = 0$.Expanding along the third column: $(1+a^2) \begin{vmatrix} 1+a^2 & -(1+a^2) \ a^2-1 & -2 \end{vmatrix} = 0$.$(1+a^2) [-2(1+a^2) + (1+a^2)(a^2-1)] = 0$.$(1+a^2)^2 [-2 + a^2 - 1] = 0$.$(1+a^2)^2 (a^2 - 3) = 0$.Since $a$ is real,$1+a^2 
eq 0$,so $a^2 - 3 = 0$,which gives $a = \pm \sqrt{3}$.Thus,$S = \{-\sqrt{3}, \sqrt{3}\}$.

If $S$ is the set of all real values of $a$ such that a plane passing through the points $(-a^2, 1, 1), (1, -a^2, 1), (1, 1, -a^2)$ also passes through the point $(-1, -1, 1)$,then $S=$

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