Probability of Union and Intersection of Events

A, B and C are three independent events such that, $$P(A) = \frac12, P(B) = \frac13, P(C) = \frac17$$

Find $$P((A\cup B)\cap C)$$.

$$P((A\cup B)\cap C) = P((A\cup B))\cdot P(C) = \frac23 \cdot \frac17 =\frac2{21}$$

But:

$$P((A\cup B)\cap C) =P((A\cap C)\cup (B\cap C)) = \frac1{14} + \frac1{21} - \frac1{14}\cdot \frac1{21} = \frac{34}{14\cdot 21}$$

Why are the two methods giving different answers?

$$A \cap C$$ and $$B \cap C$$ are not independent.
We have $$P(A \cap B \cap C)=\frac1{42}$$.
Hence, the second computation should be $$\frac1{14}+\frac1{21}-\frac1{42}=\frac1{14}+\frac1{42}=\frac{4}{42}=\frac{2}{21}$$