I swear I'm sleeping

$\lbrace a_{n} \rbrace$ has the characteristic polynomial $x^{2}-5x+6 = (x-2)(x-3)$ with, clearly, zeroes $x = 2,3$ . As such, we have a closed form for $a_{n} = c_{1}2^{n}+c_{2}3^{n}$ , where $c_{1},c_{2}$ are constants.

We then have $a_{1} = 2c_{1}+3c_{2} = 8$ and $a_{2} = 4c_{1}+9c_{2} = 21$ ; solving the system of equations gives us $c_{1} = \frac{3}{2}$ and $c_{2} = \frac{5}{3}$ . As such, we have the explicit formula

$a_{n} = 3 \cdot 2^{n-1} + 5 \cdot 3^{n-1}$

Thus, using Fermat's little theorem, we get

$a_{6001} \equiv 3 \cdot 2^{6000} + 5 \cdot 3^{6000} \equiv 3 \cdot 1^{1000} + 5 \cdot 1^{1000} \equiv \boxed{1} \pmod{7}$