If A equiv (1, -1, 0),B equiv (0, 1, -1),and | vedclass

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(C) Let $\overline{d} = p\hat{i} + q\hat{j} + r\hat{k}$,where $p, q, r \in \mathbb{R}$.Since $\overline{b}, \overline{c}, \overline{d}$ are coplanar,t

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$+\frac{1}{\sqrt{6}}(1, 1, -2)$

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(C) Let $\overline{d} = p\hat{i} + q\hat{j} + r\hat{k}$,where $p, q, r \in \mathbb{R}$.Since $\overline{b}, \overline{c}, \overline{d}$ are coplanar,their scalar triple product is zero:$\begin{vmatrix} 0 & 1 & -1 \ -1 & 0 & 1 \ p & q & r \end{vmatrix} = 0$Expanding the determinant: $0(0 - q) - 1(-r - p) + (-1)(-q - 0) = 0$$r + p + q = 0 \implies p + q + r = 0 \dots (i)$Given that $\overline{a}$ and $\overline{d}$ are perpendicular,their dot product is zero:$\overline{a} \cdot \overline{d} = (1, -1, 0) \cdot (p, q, r) = p - q = 0 \implies p = q \dots (ii)$Substituting $(ii)$ into $(i)$: $p + p + r = 0 \implies r = -2p$.Thus,$\overline{d} = p\hat{i} + p\hat{j} - 2p\hat{k} = p(\hat{i} + \hat{j} - 2\hat{k})$.Since $\overline{d}$ is a unit vector,$|\overline{d}| = 1$:$|p|\sqrt{1^2 + 1^2 + (-2)^2} = 1 \implies |p|\sqrt{6} = 1 \implies p = \pm \frac{1}{\sqrt{6}}$.For $p = \frac{1}{\sqrt{6}}$,$\overline{d} = \frac{1}{\sqrt{6}}(1, 1, -2)$,which matches option $(C)$.

If $A \equiv (1, -1, 0)$,$B \equiv (0, 1, -1)$,and $C \equiv (-1, 0, 1)$,then the unit vector $\overline{d}$ such that $\overline{a}$ and $\overline{d}$ are perpendicular and $\overline{b}, \overline{c}, \overline{d}$ are coplanar is

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