Numbers will be numbers

Notice that R.H.S. is divisible by 3, then from the left side we have:

$a + 2b \equiv 0 \mod 3$

we would arrive at $a \equiv b \mod 3$

Case 1 : $a \equiv b \equiv 0 \mod 3$ , let $a = 3m$ and $b = 3n$

$3m + 6n + 3c = 69$

$m + 2n + c = 23$

Case 1.1 : $m$ and $c$ are even and odd respectively, here, we can let $m = 2m'$ and $c = 2c' + 1$ .

$2m' + 2n + 2c' = 22$

$m' + n + c' = 11$ , now by stars and bars , we should have the number of solutions to be ${11 + 2}\choose{11}$ which is 78.

Case 1.2 : $m$ and $c$ are odd and even respectively, here, we can let $m = 2m' + 1$ and $c = 2c'$

This yields the same result as in Case 1.1, therefore we also have 78 solutions here.

Case 2 : $a \equiv b \equiv 1 \mod 3$ , let $a = 3m + 1$ and $b = 3n +1$ .

$3m + 6n + 3c = 66$

$m + 2n + c = 22$

Case 2.1 : $m$ and $c$ are both odd. Let $m = 2m' + 1$ and $c = 2c' + 1$ .

$2m' + 2n + 2c' = 20$

$m' + n + c' = 10$ , again by Stars and Bars, we have ${10 + 2}\choose{10}$ or 66 solutions

Case 2.2 : $m$ and $c$ are both even. Let $m = 2m'$ and $c = 2c'$ .

We arrive at $m' + n + c' = 11$ , which we already got (78 solutions)

Case 3 : $a \equiv b \equiv 2 \mod 3$ , let $a = 3m + 2$ and $b = 3n +2$ .

$3m + 6n + 3c = 63$

$m + 2n + c = 21$

Case 3.1 : $m$ and $c$ are even and odd respectively, here, we can let $m = 2m'$ and $c = 2c' + 1$ .

$2m' + 2n + 2c' = 20$

$m' + n + c' = 10$ (66 solutions)

Case 3.2 : $m$ and $c$ are odd and even respectively, here, we can let $m = 2m' + 1$ and $c = 2c'$ .

$2m' + 2n + 2c' = 20 \Rightarrow m' + n + c' = 10$ which has 66 solutions

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Therefore, we have a total of $78 + 78 + 66 + 78 + 66 + 66 = \boxed{432}$ solutions.