Major product formed when 1,1,1-trichloropropane is treated with aqueous KOH
Figure: 1,1,1-trichloropropane     

${\color{green}{\checkmark}}$ The above compound involves $\color{red}{\ce{C-H, C-C, C-Cl}}$ bonds, whose bond energies are $413, 348, 328 kj/mol $ respectively. So, initially $\color{orange}{\ce{C-Cl}}$ bond must break.  

${\color{green}{\checkmark}}$ It may be noted that generally alcoholic $\color{blue}{\ce{KOH}}$ causes elimination (dehydrohalogenation reaction) in the molecule of haloalkane, while aqueous $\color{blue}{\ce{KOH}}$ leads to substitution. As in your reaction, aqueous $\color{blue}{\ce{KOH}}$ is used, substitution reaction is most preferable.  

${\color{green}{\checkmark}}$ As primary carbon and strong $\color{orange}{\ce{C-Cl}}$ (compared to other $\ce{C-Br, C-I}$ bond) is involved in 1,1,1-trichloropropane, $S_{N2}$ mechanism is preferable.  

From the above thought process, the most preferable reaction would be:
$$\ce{Cl_3C-CH_2-CH_3 + 3KOH(aq) -&gt;[\text{heat}] (OH)3C-CH_2-CH_3 + 3KCl} $$  

${\color{green}{\checkmark}}$ It may be noted that more than one $\color{blue}{OH}$ group cannot be present on the same carbon atom. In such a case, the compound will be extremely unstable and will readily lose a water molecule to form more stable aldehyde, ketone or carboxylic acid.  

$$\ce{(OH)_3C-CH_2-CH_3 -&gt; (OH)CO-CH_2-CH_3 + H_2O}$$    

You are right, propionic acid along with water molecule is going to be the product.