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If the vectors $\bar{a}, \bar{b}, \bar{c}$ are non coplanar, then $\frac{[\bar{a}+2 \bar{b} \quad \bar{b}+2 \bar{c} \quad \bar{c}+2 \bar{a}]}{[\bar{a} \bar{b} \bar{c}]}=$
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The correct answer is:
9
Here
$\left[\begin{array}{ccc}\bar{a}+2 \bar{b} & \bar{b}+2 \bar{c} & \bar{c}+2 \bar{a}\end{array}\right]$
$\begin{aligned}[\bar{a}&+2 \bar{b} \quad \bar{b}+2 \bar{c} \quad \bar{c}+2 \bar{a}] \\ &=(\bar{a}+2 \bar{b}) \cdot[(\bar{b}+2 \bar{c}) \times(\bar{c}+2 \bar{a})] \\ &=(\bar{a}+2 \bar{b}) \cdot[(\bar{b} \times \bar{c})+(2 \bar{b} \times \bar{a})+(2 \bar{c} \times \bar{c})+(4 \bar{c} \times \bar{a})] \\ &=(\bar{a}+2 \bar{b}) \cdot[(\bar{b} \times \bar{c})+2(\bar{b} \times \bar{a})+0+4(\bar{c} \times \bar{a})] \\ &=[\bar{a}(\bar{b} \times \bar{c})]+0+0+0+0+8[\bar{b}(\bar{c} \times \bar{a})] \\ &=9[\bar{a}(\bar{b} \times \bar{c})] \\ &=9\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right] \end{aligned}$
Now
$$
\frac{\left[\begin{array}{lll}
\bar{a}+2 \bar{b} & \bar{b}+2 \bar{c} & \bar{c}+2 \bar{a}
\end{array}\right]}{\left[\begin{array}{lll}
\bar{a} & \bar{b} & \bar{c}
\end{array}\right]}=\frac{9[\bar{a} \bar{b} \bar{c}]}{[\bar{a} \bar{b} \bar{c}]}=9
$$
$\left[\begin{array}{ccc}\bar{a}+2 \bar{b} & \bar{b}+2 \bar{c} & \bar{c}+2 \bar{a}\end{array}\right]$
$\begin{aligned}[\bar{a}&+2 \bar{b} \quad \bar{b}+2 \bar{c} \quad \bar{c}+2 \bar{a}] \\ &=(\bar{a}+2 \bar{b}) \cdot[(\bar{b}+2 \bar{c}) \times(\bar{c}+2 \bar{a})] \\ &=(\bar{a}+2 \bar{b}) \cdot[(\bar{b} \times \bar{c})+(2 \bar{b} \times \bar{a})+(2 \bar{c} \times \bar{c})+(4 \bar{c} \times \bar{a})] \\ &=(\bar{a}+2 \bar{b}) \cdot[(\bar{b} \times \bar{c})+2(\bar{b} \times \bar{a})+0+4(\bar{c} \times \bar{a})] \\ &=[\bar{a}(\bar{b} \times \bar{c})]+0+0+0+0+8[\bar{b}(\bar{c} \times \bar{a})] \\ &=9[\bar{a}(\bar{b} \times \bar{c})] \\ &=9\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right] \end{aligned}$
Now
$$
\frac{\left[\begin{array}{lll}
\bar{a}+2 \bar{b} & \bar{b}+2 \bar{c} & \bar{c}+2 \bar{a}
\end{array}\right]}{\left[\begin{array}{lll}
\bar{a} & \bar{b} & \bar{c}
\end{array}\right]}=\frac{9[\bar{a} \bar{b} \bar{c}]}{[\bar{a} \bar{b} \bar{c}]}=9
$$
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