Answer
We have proven parts (e),(d) of the theorem.
Work Step by Step
Taking into the account the property
$k\left|\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right|=\left|\begin{array}{ccc}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ k a_{31} & k a_{32} & k a_{33}\end{array}\right|=\left|\begin{array}{ccc}a_{11} & a_{12} & a_{13} \\ k a_{21} & k a_{22} & k a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right|$
We get that
\[
\begin{aligned}
(\vec{u} \times \vec{v})k=\left|\begin{array}{ccc}
\hat{1} & \hat{j} & \hat{k} \\
u_{1} & u_{2} & u_{3} \\
v_{1} & v_{2} & v_{3}
\end{array}\right|k &=\left|\begin{array}{ccc}
\hat{\imath} & \hat{\jmath} & \hat{k} \\
k u_{1} & k u_{2} & k u_{3} \\
v_{1} & v_{2} & v_{3}
\end{array}\right| \\
&=\left\langle k u_{1}, k u_{2}, k u_{3}\right\rangle \times\left\langle v_{1}, v_{2}, v_{3}\right\rangle \\
&=(k \vec{u}) \times \vec{v}
\end{aligned}
\]
Similarly,
\[
\begin{aligned}
(\vec{u} \times \vec{v})\dot{k} &=\left|\begin{array}{ccc}
\hat{\imath} & \hat{\jmath} & k \\
u_{1} & u_{2} & u_{3} \\
k v_{1} & k v_{2} & k v_{3}
\end{array}\right| \\
&=\left\langle u_{1}, u_{2}, u_{3}\right\rangle \times\left\langle k v_{1}, k v_{2}, k v_{3}\right\rangle \\
&=\vec{u} \times (k \vec{v})
\end{aligned}
\]
Part e) from the properties:
\[
i) \ \overrightarrow{0}=0(\vec{v})
\]
$$ii) \ -\vec{v} \times \vec{u}=\vec{u} \times \vec{v}$$
\[
\text { iii) } (\vec{u} \times \vec{v})k=\vec{u} \times(k \vec{v})
\]
For $0=k$:
\[
\begin{array}{l}
0(\vec{u} \times \vec{v})=\vec{u} \times(0 \vec{v})=\vec{u} \times \overrightarrow{0}=\overrightarrow{0} \\
\vec{u} \times \overrightarrow{0}=-\overrightarrow{0} \times \vec{u}=\overrightarrow{0}
\end{array}
\]
Finally, we have that
\[
\vec{u} \times \overrightarrow{0}=\overrightarrow{0} \times \vec{u}=\overrightarrow{0}
\]
