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(D) We know that the scalar triple product $[\overline{a} \overline{b} \overline{c}] = (\overline{a} 	imes \overline{b}) \cdot \overline{c}$.For the

Vedclass · Accepted Answer

$3$

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(D) We know that the scalar triple product $[\overline{a} \overline{b} \overline{c}] = (\overline{a} 	imes \overline{b}) \cdot \overline{c}$.For the first term: $\overline{p} \cdot (\overline{a} + \overline{b}) = \overline{p} \cdot \overline{a} + \overline{p} \cdot \overline{b}$.Substituting $\overline{p} = \frac{\overline{b} 	imes \overline{c}}{[\overline{a} \overline{b} \overline{c}]}$,we get:$\overline{p} \cdot \overline{a} = \frac{(\overline{b} 	imes \overline{c}) \cdot \overline{a}}{[\overline{a} \overline{b} \overline{c}]} = \frac{[\overline{b} \overline{c} \overline{a}]}{[\overline{a} \overline{b} \overline{c}]} = \frac{[\overline{a} \overline{b} \overline{c}]}{[\overline{a} \overline{b} \overline{c}]} = 1$.$\overline{p} \cdot \overline{b} = \frac{(\overline{b} 	imes \overline{c}) \cdot \overline{b}}{[\overline{a} \overline{b} \overline{c}]} = 0$ (since $\overline{b} 	imes \overline{c}$ is perpendicular to $\overline{b}$).Thus,$\overline{p} \cdot (\overline{a} + \overline{b}) = 1 + 0 = 1$.Similarly,$\overline{q} \cdot (\overline{b} + \overline{c}) = 1$ and $\overline{r} \cdot (\overline{c} + \overline{a}) = 1$.Adding these,we get $1 + 1 + 1 = 3$.

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