$\sum_{i=1}^{18}\left(x_{i}-\alpha\right)=36, \sum_{i=1}^{18}\left(x_{i}-\beta\right)^{2}=90$

$\Rightarrow \sum_{i=1}^{18} x_{i}=18(\alpha+2), \sum_{i=1}^{18} x_{i}^{2}-2 \beta \sum_{i=1}^{18} x_{i}+18 \beta^{2}=90$

Hence $\sum x _{ i }^{2}=90-18 \beta^{2}+36 \beta(\alpha+2)$

Given $\frac{\sum x _{ i }^{2}}{18}-\left(\frac{\sum x _{ i }}{18}\right)^{2}=1$

$\Rightarrow 90-18 \beta^{2}+36 \beta(\alpha+2)-18(\alpha+2)^{2}=18$

$\Rightarrow 5-\beta^{2}+2 \alpha \beta+4 \beta-\alpha^{2}-4 \alpha-4=1$

$\Rightarrow(\alpha-\beta)^{2}+4(\alpha-\beta)=0 \Rightarrow|\alpha-\beta|=0$ or $4$

As $\alpha$ and $\beta$ are distinct $|\alpha-\beta|=4$