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The value of $[a-b \quad b-c \quad c-a]$, where $|a|=1$, $|\mathrm{b}|=5|\mathrm{c}|=3$, is
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$[\mathbf{a}-\mathbf{b} \mathbf{b}-\mathbf{c} \mathbf{c}-\mathbf{a}]=(\mathbf{a}-\mathbf{b}) \cdot[(\mathbf{b}-\mathbf{c}) \times(\mathbf{c}-\mathbf{a})]$
$=(\mathbf{a}-\mathbf{b}) \cdot[\mathbf{b} \times \mathbf{c}-\mathbf{b} \times \mathbf{a}-\mathbf{c} \times \mathbf{c}+\mathbf{c} \times \mathbf{a})]$
$=(\mathbf{a}-\mathbf{b}) .[\mathbf{b} \times \mathbf{c}-\mathbf{b} \times \mathbf{a}+\mathbf{c} \times \mathbf{a}]$
$=\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})-\mathbf{a} \cdot(\mathbf{b} \times \mathbf{a})+\mathbf{a} \cdot(\mathbf{c} \times \mathbf{a})-\mathbf{b}(\mathbf{b} \times \mathbf{c})$
$+\mathbf{b} \cdot(\mathbf{b} \times \mathbf{a})-\mathbf{b}(\mathbf{c} \times \mathbf{a})$
$=[\mathbf{a b c}]-[\mathbf{a b a}]+[\mathbf{a c a}]-[\mathbf{b b c}]+[\mathbf{b b a}]-[$ bca $]$
$=[\mathbf{a b c}]-[\mathbf{b c a}]=0 \quad\{\because[\mathbf{a b c}]=[\mathbf{b c a}]=[\mathbf{c a b}]\}$
$=(\mathbf{a}-\mathbf{b}) \cdot[\mathbf{b} \times \mathbf{c}-\mathbf{b} \times \mathbf{a}-\mathbf{c} \times \mathbf{c}+\mathbf{c} \times \mathbf{a})]$
$=(\mathbf{a}-\mathbf{b}) .[\mathbf{b} \times \mathbf{c}-\mathbf{b} \times \mathbf{a}+\mathbf{c} \times \mathbf{a}]$
$=\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})-\mathbf{a} \cdot(\mathbf{b} \times \mathbf{a})+\mathbf{a} \cdot(\mathbf{c} \times \mathbf{a})-\mathbf{b}(\mathbf{b} \times \mathbf{c})$
$+\mathbf{b} \cdot(\mathbf{b} \times \mathbf{a})-\mathbf{b}(\mathbf{c} \times \mathbf{a})$
$=[\mathbf{a b c}]-[\mathbf{a b a}]+[\mathbf{a c a}]-[\mathbf{b b c}]+[\mathbf{b b a}]-[$ bca $]$
$=[\mathbf{a b c}]-[\mathbf{b c a}]=0 \quad\{\because[\mathbf{a b c}]=[\mathbf{b c a}]=[\mathbf{c a b}]\}$
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