$\times:\mathbb{R}^3\times\mathbb{R}^3\mapsto\mathbb{R}^3$ 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.