2-Chloro-2-methylpentane on reaction with sodium methoxide yields

 

Strong base like ${\mathrm{CH}}_3{\mathrm{O}}^{-}$ leads to more substitution.

Sterically hindrance around the reacting carbon leads to more elimination. In elimination, major product is of more substituted alkene (Saytzeff rule). In the given reaction ${\mathrm{CH}}_3{\mathrm{O}}^{-}$ is unhindered whereas reacting carbon of 2-chloro-2-methylpentane is sterically hindered and thus all the three given compounds will be formed.