Solution
\[
\begin{aligned}
& \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=2 \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}} \Rightarrow \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}} \times(2 \overrightarrow{\mathrm{c}})=0 \\
& \Rightarrow \overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}}-2 \overrightarrow{\mathrm{c}})=0
\end{aligned}
\]
$\vec{a} \& \vec{b}-2 \vec{c}$ are parallel
\[
\overrightarrow{\mathrm{b}}-2 \overrightarrow{\mathrm{c}}=\lambda \overrightarrow{\mathrm{a}}
\]
Square $(\overrightarrow{\mathrm{b}}-2 \overrightarrow{\mathrm{c}})^{2}=(\lambda \overrightarrow{\mathrm{a}})^{2}$
\[
\begin{aligned}
& \Rightarrow|\vec{b}|^{2}+4|\vec{c}|^{2}-4 \vec{b} \cdot \vec{c}=\lambda^{2}|\vec{a}|^{2} \\
& \Rightarrow 16+4(4)-4|\vec{b} \| \vec{c}| \cos 60^{\circ}=\lambda^{2}(1) \\
& \Rightarrow 32-4(4)(2)\left(\frac{1}{2}\right)=\lambda^{2} \Rightarrow \lambda^{2}=16 \Rightarrow \lambda= \pm 4 \\
& \because \vec{b}-2 \vec{c}=\lambda \vec{a} \operatorname{dot} \text { with } \vec{c} \\
& \vec{b} \cdot \vec{c}-2 \vec{c} \cdot \vec{c}=\lambda \vec{a} \cdot \vec{c} \\
& \Rightarrow|\vec{b} \| \vec{c}| \cos 60^{\circ}-2|\vec{c}|^{2}=\lambda(\vec{a} \cdot \vec{c}) \\
& \Rightarrow 4(2)\left(\frac{1}{2}\right)-2(4)=\lambda \vec{a} \cdot \vec{c} \Rightarrow \lambda(\vec{a} \cdot \vec{c})=-4 \\
& \Rightarrow \lambda \| \vec{a} \cdot \vec{c}|=4 \Rightarrow| \vec{a} \cdot \vec{c} \mid=1 \operatorname{Answer}(4)
\end{aligned}
\]
.
