Consider three vectors $\vec{a}, \vec{b}, \vec{c}$. Let $|\vec{a}|=2,|\vec{b}|=3$ and $\vec{a}=\vec{b} \times \vec{c}$. If $\alpha \in\left[0, \frac{\pi}{3}\right]$ is the angle between the vectors $\vec{b}$ and $\vec{c}$, then the minimum value of $27|\vec{c}-\vec{a}|^2$ is equal to:

<p>$$\begin{aligned} & \vec{a}=\vec{b} \times \vec{c} \\ & |\vec{a}|=2,|\vec{b}|=3 \end{aligned}$$</p> <p>$\vec{a} \cdot \vec{b}=0$ and $\vec{a} \cdot \vec{c}=0$</p> <p>$$\begin{aligned} & |\vec{c}-\vec{a}|^2=|\vec{c}|^2+|\vec{a}|^2-2 \vec{c} \cdot \vec{a} \\ & =4+|\vec{c}|^2 \end{aligned}$$</p> <p>$$\begin{aligned} & |\vec{a}|=|\vec{b} \times \vec{c}|=|\vec{b}| \sin \alpha|\vec{c}| \\ & \Rightarrow \sin \alpha|\vec{c}|=\frac{2}{3} \\ & \Rightarrow \sin ^2 \alpha=\frac{4}{9|\vec{c}|^2} \\ & \Rightarrow|\vec{c}|^2=\frac{4}{9 \sin ^2 \alpha} \\ & \Rightarrow|\vec{c}-\vec{a}|^2=4+\frac{4}{9 \sin ^2 \alpha} \end{aligned}$$</p> <p>For $|\vec{c}-\vec{a}|^2$ to be minimum for $\alpha \in\left[0, \frac{\pi}{3}\right]$</p> <p>$$\begin{gathered} \sin \alpha=\frac{\sqrt{3}}{2} \\ 27|\vec{c}-\vec{a}|^2=27\left[4+\frac{4.4}{9.3}\right] \\ =27\left[\frac{124}{27}\right]=124 \end{gathered}$$</p>