$$\cfrac { 1 }{ 2 } +\cfrac { 2 }{ 3 } +\cfrac { 3 }{ 4 } =\cfrac{23}{12}=1\cfrac{11}{12}$$
add them if the denominators are common. Now as LCD of the denominators $$2,3$$ and $$4$$ is $$12$$
$$\cfrac { 1 }{ 2 } +\cfrac { 2 }{ 3 } +\cfrac { 3 }{ 4 } =(\cfrac { 1 }{ 2 } \times 1)+(\cfrac { 2 }{ 3 } \times 1)+(\cfrac { 3 }{ 4 } \times 1)$$
$$=\cfrac{1\times 6}{2\times 6}+\cfrac{2\times 4}{3\times 4}+\cfrac{3\times 3}{4\times 3}$$
$$=\cfrac{6+8+9}{12}=\cfrac{23}{12}$$