$\times:\mathbb{R}^3\times\mathbb{R}^3\mapsto\mathbb{R}^3$ is an antisymmetric bilinear form. This means that
$$
\left(a\vec{x}+b\vec{y}\right)\times\vec{z}=a\left(\vec{x}\times\vec{z}\right)+b\left(\vec{y}\times\vec{z}\right)\tag{1}
$$
and
$$
\vec{y}\times\vec{x}=-\vec{x}\times\vec{y}\tag{2}
$$
It is not associative; that is, one cannot claim that $\left(\vec{x}\times\vec{y}\right)\times\vec{z}=\vec{x}\times\left(\vec{y}\times\vec{z}\right)$.
An immediate consequence of $(2)$ is that
$$
x\times x=0\tag{3}
$$
because $x\times x=-x\times x$.
A consequence of $(1)$ and $(2)$ is that
$$
\begin{align}
\vec{z}\times\left(a\vec{x}+b\vec{y}\right)
&=-\left(a\vec{x}+b\vec{y}\right)\times\vec{z}\\
&=-a\left(\vec{x}\times\vec{z}\right)-b\left(\vec{y}\times\vec{z}\right)\\
&=a\left(\vec{z}\times\vec{x}\right)+b\left(\vec{z}\times\vec{y}\right)\tag{4}
\end{align}
$$
On the canonical basis for $\mathbb{R}^3$, $\{i,j,k\}$, $\times$ is defined by $(1)$, $(2)$, $(3)$, and
$$
\begin{align}
i\times j&=k\\
j\times k&=i\\
k\times i&=j
\end{align}\tag{5}
$$
Hopefully, applying these properties to your problems should help.