If$$\frac{1}{x+2}=\frac{3}{y+3}=\frac{1331}{z+1331}=\frac{1}{3}$$, then what is the value of $$\frac{x}{x+1}+\frac{4}{y+2}+\frac{z}{z+2662}?$$

Given $$\frac{1}{x+2}=\frac{1}{3}$$

We get x = 1 .....(1)

$$\frac{3}{y+3}=\frac{1}{3}$$

we get y = 6 .....(2)

$$\frac{1331}{z+1331}=\frac{1}{3}$$

we get z = 2662 ....(3)

We need to find $$\frac{x}{x+1}+\frac{3}{y+3}+\frac{z}{z+2662}$$

Substitute equations (1), (2) and (3) in the above equation

= $$\frac{1}{1+1}+\frac{3}{6+3}+\frac{2662}{2662+2662}$$

= $$\frac{1}{2}+\frac{3}{9}+\frac{1}{2}$$

= $$\frac{3}{2}$$

Hence, option C is the correct answer.