I decided to learn differential geometry by myself, and I am confused with the definition of the multiplication of 1-forms. According to the book that I am currently reading [1], given two 1-forms $\omega$ and $\nu$ on $T\mathbb{R}^2$, and vectors $V_1,V_2\in\mathbb{R}^2$, the operation $$ \omega\wedge\nu(V_1,V_2)=\begin{vmatrix}\omega(V_1)&\nu(V_1)\\ \omega(V_2)&\nu(V_2)\end{vmatrix}$$ defines the multiplication of the 1-forms $\omega$ and $\nu$.
In Exercise 3.14 of [1] the author states
Show that, for any 1-forms $\omega$ and $\nu$ on $T\mathbb{R}^3$, there exists constants $c_1$, $c_2$ and $c_3$ such that $$ \omega\wedge\nu=c_1 dx\wedge dy+c_2 dx\wedge dz + c_3 dy\wedge dz .$$
Why is this true?
[1] D. Bachman "A geometric approach to differential forms", Birkhauser, 2nd ed. 2012.
Let $\omega = a dx + b dy + c dz$ and $\nu = p dx + q dy + r dz$. Then just compute using the determinant formula you gave:
$$ \begin {align*} \omega \wedge \nu (V,W) &= \left| \begin{array}{cc} \omega(V) & \nu(V) \\ \omega(W) & \nu(W) \end{array} \right| \\ &= \left| \begin{array}{cc} av_1+bv_2+cv_3 & pv_1+qv_2+rv_3 \\ aw_1+bw_2+cw_3 & pw_1+qw_2+rw_3 \end{array} \right| \\ &= (av_1+b_2+cv_3)(pw_1+qw_2+rw_3) - (pv_1+qv_2+rv_3)(aw_1+bw_2+cw_3) \end {align*} $$
After you multiply it out, some terms cancel, and what's left factors as:
$$ (aq-bp)(v_1w_2-v_2w_1) + (ar-cp)(v_1w_3-v_3w_1) + (br-cq)(v_2w_3-v_3w_2) $$
Notice that $v_1w_2-v_2w_1 = dx \wedge dy(V,W)$, and similarly $v_1w_3-v_3w_1 = dx \wedge dz(V,W)$ and $v_2w_3-v_3w_2 = dy \wedge dz(V,W)$.