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$3^{2002} = 9^{1001}$ and $7^{2002} = 49^{1001}$ .

$a^{n} + b^{n} = (a + b)(a^{n-1} - a^{n-2}\times{b} + a^{n-3}\times{b^{2}} - \cdot \cdot \cdot - b^{n-1})$ where $n$ is odd.

$\therefore (a + b) \mid (a^{n} + b^{n}$ ) , where $n$ is odd.

$\therefore (9 + 49 = 58) \mid 9^{1001} + 49^{1001} \Rightarrow 29 \mid 9^{1001} + 49^{1001}$ .

$\Rightarrow 3^{2002} + 7^{2002} + 2002\equiv 2002\equiv1\pmod{29}$

$\begin{aligned} 3^{2002}+7^{2003}+2002 & \pmod {29} \\ 3^{71\phi(29)}\cdot 3^{14}+7^{ 71\phi(29)}\cdot 7^{14}+29(69)+1 & \pmod {29} \\ 3^{14}+7^{14}+1 \equiv (3^{3})^4 3^2+(7^{3})^4 7^2+1 & \pmod {29} \\ (27)^4 9+(343)^4 49+1 \equiv (-2)^4 9+(-5)^4 20+1 & \pmod {29} \\ 16\cdot 9 + (25)^2 20+1 \equiv 144+(-4)^2 20+1 & \pmod {29} \\ 28+320+1 = 349 \equiv \boxed{1} & \pmod {29} \end{aligned}$