If vec{a}, vec{b}, vec{c} are three non-copla | vedclass

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(D) Let $V = [\vec{a} \vec{b} \vec{c}]$. By definition, $\vec{a} \cdot \vec{p} = \vec{a} \cdot \frac{\vec{b} 	imes \vec{c}}{V} = \frac{[\vec{a} \vec{

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$3$

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(D) Let $V = [\vec{a} \vec{b} \vec{c}]$. By definition, $\vec{a} \cdot \vec{p} = \vec{a} \cdot \frac{\vec{b} 	imes \vec{c}}{V} = \frac{[\vec{a} \vec{b} \vec{c}]}{V} = 1$.Similarly, $\vec{b} \cdot \vec{q} = 1$ and $\vec{c} \cdot \vec{r} = 1$.Also, $\vec{b} \cdot \vec{p} = 0$, $\vec{c} \cdot \vec{p} = 0$, $\vec{c} \cdot \vec{q} = 0$, $\vec{a} \cdot \vec{q} = 0$, $\vec{a} \cdot \vec{r} = 0$, and $\vec{b} \cdot \vec{r} = 0$ because the dot product of a vector with a cross product containing itself is zero.Now, expand the expression: $(\vec{a} + \vec{b}) \cdot \vec{p} + (\vec{b} + \vec{c}) \cdot \vec{q} + (\vec{c} + \vec{a}) \cdot \vec{r} = (\vec{a} \cdot \vec{p} + \vec{b} \cdot \vec{p}) + (\vec{b} \cdot \vec{q} + \vec{c} \cdot \vec{q}) + (\vec{c} \cdot \vec{r} + \vec{a} \cdot \vec{r})$.Substituting the values: $(1 + 0) + (1 + 0) + (1 + 0) = 3$.

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