Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three unit vectors such that $|\vec{a}-\vec{b}|^{2}+|\vec{a}-\vec{c}|^{2}=8$
Then $|\vec{a}+2 \vec{b}|^{2}+|\vec{a}+2 \vec{c}|^{2}$ is equal to
$|\vec{a}|=|\vec{b}|=|\vec{c}|=1$
$|\vec{a}-\vec{b}|^{2}+|\vec{a}-\vec{b}|^{2}=8$
$\Rightarrow|\vec{a}|^{2}+|\vec{b}|^{2}-2 \vec{a} \cdot \vec{b}+|\vec{a}|^{2}+|\vec{c}|^{2}-2 \vec{a} \cdot \vec{c}=8$
$\Rightarrow \quad 4-2(\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c})=8$
$\Rightarrow \vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}=-2$
$|\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}|^{2}+|\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{c}}|^{2}$
$=\left|a^{2}\right|+4|\vec{b}|^{2}+4 \vec{a} \cdot \vec{b}+|\vec{a}|^{2}+4|\vec{c}|^{2}+4 \vec{a} \cdot \vec{c}$
$=10+4(\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c})$
$=10-8$
$=2$
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