= Solution
{c}
For <vectors> $\mathbf a,\mathbf b,\mathbf c\in\mathbb R^3$, the <scalar product>, <cross product> and <scalar triple product> are respectively
$$
\begin{aligned}
\mathbf a\cdot\mathbf b&=a_1b_1+a_2b_2+a_3b_3,\\
\mathbf a\wedge\mathbf b&=(a_2b_3-a_3b_2,\ a_3b_1-a_1b_3,\ a_1b_2-a_2b_1),\\
(\mathbf a\wedge\mathbf b)\cdot\mathbf c&=\det\begin{pmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{pmatrix}.
\end{aligned}
$$
Expanding the <dot product> gives
$$
(\mathbf a\wedge\mathbf b)\cdot\mathbf a
=a_1a_2b_3-a_1a_3b_2+a_2a_3b_1-a_1a_2b_3+a_1a_3b_2-a_2a_3b_1=0.
$$
Interchanging $\mathbf a$ and $\mathbf b$ changes the sign of their <cross product>, so its <dot product> with $\mathbf b$ is also zero. Thus the <cross product> is <orthogonal> to both <vectors>, including when it is the <zero vector>.
The displayed <scalar triple product> is a <determinant>. A square <matrix> has nonzero <determinant> exactly when its rows are <linearly independent>. Hence
$$
\boxed{(\mathbf a\wedge\mathbf b)\cdot\mathbf c\ne0\iff\mathbf a,\mathbf b,\mathbf c\text{ are linearly independent}.}
$$
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