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

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(C) For the four points to be coplanar,the determinant of the vectors formed by them must be zero. Let the points be $A(-\lambda^2, 1, 1)$,$B(1, -\lam

Vedclass · Accepted Answer

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

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(C) For the four points to be coplanar,the determinant of the vectors formed by them must be zero. Let the points be $A(-\lambda^2, 1, 1)$,$B(1, -\lambda^2, 1)$,$C(1, 1, -\lambda^2)$,and $D(-1, -1, 1)$.The vectors $\vec{AB} = (1+\lambda^2, -\lambda^2-1, 0)$,$\vec{AC} = (1+\lambda^2, 0, -\lambda^2-1)$,and $\vec{AD} = (-1+\lambda^2, -2, 0)$ must be coplanar.Alternatively,using the condition that the four points $(x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3), (x_4, y_4, z_4)$ are coplanar:$\begin{vmatrix} x_1-x_4 & y_1-y_4 & z_1-z_4 \ x_2-x_4 & y_2-y_4 & z_2-z_4 \ x_3-x_4 & y_3-y_4 & z_3-z_4 \end{vmatrix} = 0$Substituting the points:$\begin{vmatrix} -\lambda^2+1 & 2 & 0 \ 2 & -\lambda^2+1 & 0 \ 2 & 2 & -\lambda^2-1 \end{vmatrix} = 0$Expanding along the third column:$(-\lambda^2-1) [(-\lambda^2+1)^2 - 4] = 0$$(-\lambda^2-1) [(\lambda^4 - 2\lambda^2 + 1) - 4] = 0$$(-\lambda^2-1) (\lambda^4 - 2\lambda^2 - 3) = 0$$(-\lambda^2-1) (\lambda^2-3) (\lambda^2+1) = 0$Since $\lambda$ is real,$\lambda^2+1 
eq 0$. Thus,$\lambda^2-3 = 0$,which gives $\lambda^2 = 3$.Therefore,$\lambda = \pm \sqrt{3}$.So,$S = \{-\sqrt{3}, \sqrt{3}\}$.

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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