The number of distinct real values of lambda, | vedclass

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(B) For three vectors to be coplanar,their scalar triple product must be zero. Thus,we have the determinant equation:$\left|\begin{array}{ccc} -\lambd

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two.

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(B) For three vectors to be coplanar,their scalar triple product must be zero. Thus,we have the determinant equation:$\left|\begin{array}{ccc} -\lambda^2 & 1 & 1 \ 1 & -\lambda^2 & 1 \ 1 & 1 & -\lambda^2 \end{array}ight| = 0$Expanding the determinant along the first row:$-\lambda^2(\lambda^4 - 1) - 1(-\lambda^2 - 1) + 1(1 + \lambda^2) = 0$$-\lambda^6 + \lambda^2 + \lambda^2 + 1 + 1 + \lambda^2 = 0$$-\lambda^6 + 3\lambda^2 + 2 = 0$$\lambda^6 - 3\lambda^2 - 2 = 0$Let $t = \lambda^2$. Then the equation becomes $t^3 - 3t - 2 = 0$.By testing roots,for $t = -1$,$(-1)^3 - 3(-1) - 2 = -1 + 3 - 2 = 0$. So $(t+1)$ is a factor.$(t+1)(t^2 - t - 2) = 0$$(t+1)(t+1)(t-2) = 0$$(t+1)^2(t-2) = 0$Thus,$t = -1$ or $t = 2$.Since $t = \lambda^2$,we have $\lambda^2 = -1$ (no real solution) or $\lambda^2 = 2$.For real $\lambda$,$\lambda = \pm \sqrt{2}$.Therefore,there are two distinct real values of $\lambda$.

The number of distinct real values of $\lambda$,for which the vectors $-\lambda^2 \hat{i}+\hat{j}+\hat{k}$,$\hat{i}-\lambda^2 \hat{j}+\hat{k}$ and $\hat{i}+\hat{j}-\lambda^2 \hat{k}$ are coplanar,is

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