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320)

$\small \dfrac{5}{3} : \left(\dfrac{2}{3}+\dfrac{1}{6}\right) : \left(1-\dfrac{1}{6}-\dfrac{1}{4}\right) +\dfrac{2}{\cancel{11}_1}·\dfrac{\cancel{22}^2}{7}-1+\dfrac{1}{3}=$

$\small =\dfrac{5}{3} : \left(\dfrac{4+1}{6}\right) : \left(\dfrac{12-2-3}{12}\right) +\dfrac{2}{1}·\dfrac{2}{7}-1+\dfrac{1}{3}=$

$\small =\dfrac{5}{3} : \dfrac{5}{6} : \dfrac{7}{12} +\dfrac{4}{7}-1+\dfrac{1}{3}=$

$\small =\dfrac{5}{3}·\dfrac{6}{5}·\dfrac{12}{7} +\dfrac{4}{7}-1+\dfrac{1}{3}=$

$\small =\dfrac{\cancel5^1}{\cancel3_1}·\dfrac{\cancel6^2}{\cancel5_1}·\dfrac{12}{7} +\dfrac{4}{7}-1+\dfrac{1}{3}=$

$\small =\dfrac{1}{1}·\dfrac{2}{1}·\dfrac{12}{7} +\dfrac{4}{7}-1+\dfrac{1}{3}=$

$\small =\dfrac{24}{7} +\dfrac{4}{7}-1+\dfrac{1}{3}=$

$\small =\dfrac{\cancel{28}^4}{\cancel7_1} -1+\dfrac{1}{3}=$

$\small = 4 -1+\dfrac{1}{3}=$

$\small = 3+\dfrac{1}{3}=$

$\small = \dfrac{9+1}{3}=$

$\small = \dfrac{10}{3}$