The sum of the distinct real values of mu,for | vedclass

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(A) Three vectors are coplanar if their scalar triple product is zero. The scalar triple product is given by the determinant of the matrix formed by t

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$-1$

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(A) Three vectors are coplanar if their scalar triple product is zero. The scalar triple product is given by the determinant of the matrix formed by the components of the vectors:$D = \begin{vmatrix} \mu & 1 & 1 \ 1 & \mu & 1 \ 1 & 1 & \mu \end{vmatrix} = 0$Expanding the determinant:$D = \mu(\mu^2 - 1) - 1(\mu - 1) + 1(1 - \mu) = 0$$D = \mu(\mu - 1)(\mu + 1) - 1(\mu - 1) - 1(\mu - 1) = 0$$D = (\mu - 1) [\mu(\mu + 1) - 1 - 1] = 0$$D = (\mu - 1)(\mu^2 + \mu - 2) = 0$$D = (\mu - 1)(\mu + 2)(\mu - 1) = 0$$D = (\mu - 1)^2(\mu + 2) = 0$The distinct real values of $\mu$ are $1$ and $-2$.The sum of these distinct real values is $1 + (-2) = -1$.

The sum of the distinct real values of $\mu$,for which the vectors $\mu \hat{i} + \hat{j} + \hat{k}$,$\hat{i} + \mu \hat{j} + \hat{k}$,and $\hat{i} + \hat{j} + \mu \hat{k}$ are coplanar,is

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