2. Given vectors $\boldsymbol{a}=(0,1), \boldsymbol{b}=\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right), \boldsymbol{c}=\left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right)$. Real numbers $x, y, z$ satisfy $x \boldsymbol{a}+y \boldsymbol{b}+z \boldsymbol{c}=(1,2)$, then the minimum value of $x^{2}+y^{2}+z^{2}$ is $\qquad$.
2. $\frac{10}{3}$.

From $x \boldsymbol{a}+y \boldsymbol{b}+z \boldsymbol{c}=x(0,1)+y\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)+z\left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right)=\left(-\frac{\sqrt{3}}{2} y+\frac{\sqrt{3}}{2} z, x-\frac{1}{2} y-\frac{1}{2} z\right)$,
we have
$$
\left\{\begin{array} { c } 
{ - \frac { \sqrt { 3 } } { 2 } y + \frac { \sqrt { 3 } } { 2 } z = 1 , } \\
{ x - \frac { 1 } { 2 } y - \frac { 1 } { 2 } z = 2 }
\end{array} \Rightarrow \left\{\begin{array}{c}
\sqrt{3} z=1-\sqrt{3}(2-x) \\
-\sqrt{3} y=1+\sqrt{3}(2-x)
\end{array}\right.\right.
$$

From this, we can get $3 x^{2}+3 y^{2}+3 z^{2}=3 x^{2}+2\left(1+3(2-x)^{2}\right)=9 x^{2}-24 x+26=9\left(x-\frac{4}{3}\right)^{2}+10$, therefore, the minimum value of $x^{2}+y^{2}+z^{2}$ is $\frac{10}{3}$, which is achieved when $x=\frac{4}{3}$.