= Solution
{c}
Write $\times$ for the paper's $\wedge$. In a right-handed <Cartesian coordinate system>, the <cross product> is
$$
\boxed{\mathbf a\times\mathbf b=(a_2b_3-a_3b_2,\ a_3b_1-a_1b_3,\ a_1b_2-a_2b_1).}
$$
It is perpendicular to both <vectors>, has <norm> $|\mathbf a||\mathbf b|\sin\phi$, where $\phi$ is their <angle>, and has direction given by the <right-hand rule> when nonzero. Its <norm> is the <area> of the <parallelogram> spanned by the two <vectors>.
If $\mathbf b\times\mathbf c\ne0$, use that <parallelogram> as the base of the <parallelepiped>. Its perpendicular height is $|\mathbf a\cdot\mathbf n|$, where $\mathbf n=(\mathbf b\times\mathbf c)/|\mathbf b\times\mathbf c|$. Thus its <volume> is
$$
\boxed{\mathcal V=|\mathbf b\times\mathbf c|\,|\mathbf a\cdot\mathbf n|=|\mathbf a\cdot(\mathbf b\times\mathbf c)|.}
$$
If $\mathbf b\times\mathbf c=0$, both the <volume> and this <scalar triple product> vanish. If the <cross product> is nonzero, the <scalar triple product> vanishes exactly when $\mathbf a$ lies in the <plane> spanned by $\mathbf b,\mathbf c$. If it is zero, $\mathbf b,\mathbf c$ are <linearly dependent>, and adjoining $\mathbf a$ still spans at most a <plane>. Consequently \b[the three <vectors> are coplanar if and only if $\mathbf a\cdot(\mathbf b\times\mathbf c)=0$], including cases involving a zero <vector>.
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