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(B) Three vectors $\vec{a}, \vec{b}, 	ext{ and } \vec{c}$ are coplanar if and only if their scalar triple product is zero,i.e.,$[\vec{a} \vec{b} \vec

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is empty

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(B) Three vectors $\vec{a}, \vec{b}, 	ext{ and } \vec{c}$ are coplanar if and only if their scalar triple product is zero,i.e.,$[\vec{a} \vec{b} \vec{c}] = 0$.The scalar triple product is given by the determinant of the components:$\begin{vmatrix} \alpha & 1 & 3 \ 2 & 1 & -\alpha \ \alpha & -2 & 3 \end{vmatrix} = 0$Expanding along the first row:$\alpha(1(3) - (-2)(-\alpha)) - 1(2(3) - \alpha(-\alpha)) + 3(2(-2) - \alpha(1)) = 0$$\alpha(3 - 2\alpha) - 1(6 + \alpha^2) + 3(-4 - \alpha) = 0$$3\alpha - 2\alpha^2 - 6 - \alpha^2 - 12 - 3\alpha = 0$$-3\alpha^2 - 18 = 0$$-3(\alpha^2 + 6) = 0$$\alpha^2 + 6 = 0$Since $\alpha \in \mathbb{R}$,$\alpha^2$ must be non-negative,so $\alpha^2 + 6 \geq 6$. Thus,there is no real value of $\alpha$ that satisfies this equation.Therefore,the set $S$ is an empty set.

Let $\alpha \in \mathbb{R}$ and the three vectors $\vec{a} = \alpha \hat{i} + \hat{j} + 3\hat{k}$,$\vec{b} = 2\hat{i} + \hat{j} - \alpha \hat{k}$,and $\vec{c} = \alpha \hat{i} - 2\hat{j} + 3\hat{k}$. Then the set $S = \{ \alpha : \vec{a}, \vec{b}, \text{ and } \vec{c} \text{ are coplanar} \}$

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