Let $\overrightarrow{AB} = 2 \hat{i} + 4 \hat{j} -5 \hat{k}$
$\overrightarrow{BC} = \hat{i} + 2 \hat{j} + 3 \hat{k}$
$\therefore \overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}$
$\; \; \; \; \;= 3 \hat{i} + 6 \hat{j} - 2 \hat{k}$
$|\overrightarrow{AC} | = \sqrt{3^2 + 6^2+(-2)^2} = 7$
$\therefore$ unit vector parallel to $\overrightarrow{AC}$ is $\frac{1}{7} (3 \hat{i} + 6 \hat{j} -2 \hat{k})$