The number of distinct real values of lambda, | vedclass

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(C) Three vectors are coplanar if their scalar triple product is zero. The scalar triple product is given by the determinant: $\left|\begin{array}{ccc

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(C) Three vectors are coplanar if their scalar triple product is zero. The scalar triple product is given by the determinant: $\left|\begin{array}{ccc}-\lambda^2 & 1 & 1 \ 1 & -\lambda^2 & 1 \ 1 & 1 & -\lambda^2\end{array}ight| = 0$ 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 $x = \lambda^2$. Then $x^3 - 3x - 2 = 0$. By testing values,$x = -1$ is a root: $(-1)^3 - 3(-1) - 2 = -1 + 3 - 2 = 0$. Dividing by $(x+1)$,we get $(x+1)(x^2 - x - 2) = 0$,which factors to $(x+1)(x-2)(x+1) = 0$. So,$(x+1)^2(x-2) = 0$. Since $x = \lambda^2$,we have $(\lambda^2+1)^2(\lambda^2-2) = 0$. For real $\lambda$,$\lambda^2+1$ cannot be zero. Thus,$\lambda^2 - 2 = 0$,which gives $\lambda = \pm \sqrt{2}$. There are $2$ 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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