Parabola is given by

$x=a{y}^2+by+c$

It passes through $\left(-2,1\right)$, $\left(1,2\right)$ & $\left(-1,3\right)$

$-2=a+b+c .....\left(i\right)$

$1=4a+2b+c .....\left(ii\right)$

$-1=9a+3b+c .....\left(iii\right)$

Equation $\left(ii\right)-\left(i\right)$ & $\left(iii\right)-\left(ii\right)$ gives

$3a+b=3 .....\left(iv\right)$

$5a+b=-2 .....\left(v\right)$

Equation $\left(v\right)-\left(iv\right)$

$2a=-5$

$\Rightarrow a=\frac{-5}{2}$

Equation $\left(iv\right)$

$3\left(\frac{-5}{2}\right)+b=3$

$\Rightarrow b=\frac{21}{2}$

Equation $\left(i\right)$

$\frac{-5}{2}+\frac{21}{2}+c=-2$

$\Rightarrow c=-10$

Parabola is, $x=\frac{-5}{2}{y}^2+\frac{21}{2}y-10$

$\Rightarrow x+10=\frac{-5}{2}\left[y^2-\frac{21}{5}y\right]$

$\Rightarrow x+10=\frac{-5}{2}\left[{\left(y-\frac{21}{10}\right)}^2+\frac{441}{10}\right]$

$\Rightarrow x+10-\frac{441}{2}=\frac{-5}{2}{\left(y-\frac{21}{10}\right)}^2$

$\Rightarrow {\left(y-\frac{21}{10}\right)}^2=-4\left(\frac{5}{8}\right)\left(x-\frac{421}{2}\right)$

Vertex $\equiv \left(\frac{421}{2},\frac{21}{10}\right)$

Focus is given by $\left(-a,0\right)$ to the parabola $y^2=-4ax$.

So, focus to the parabola ${\left(y-\frac{21}{10}\right)}^2=-4\left(\frac{5}{8}\right)\left(x-\frac{421}{2}\right)$ is $\left(-\frac{5}{8}-\frac{421}{2},\frac{21}{10}\right)=\left(-\frac{1689}{8},\frac{21}{10}\right)$

So, $y$ coordinate of focus is $\frac{21}{10}$.