You wrote it yourself: the gcd is the smallest positive linear combination. Smallest positive linear combination is shorthand for smallest positive number which is a linear combination. It is true that $0$ is a linear combination of $12$ and $6$ with integer coefficients, but $0$ is not positive.
The proof is not difficult, but it is somewhat lengthy. We give full detail below.
Let $e$ be the smallest positive linear combination $as+bt$ of $a$ and $b$, where $s$ and $t$ are integers. Suppose in particular that $e=ax+by$.
Let $d=\gcd(a,b)$. Then $d$ divides $a$ and $b$, so it divides $ax+by$. Thus $d$ divides $e$, and therefore in particular $d\le e$.
We show that in fact $e$ is a common divisor of $a$ and $b$, which will imply that $e\le d$. That, together with our earlier $d\le e$, will imply that $d=e$.
So it remains to show that $e$ divides $a$ and $e$ divides $b$. We show that $e$ divides $a$. The proof that $e$ divides $b$ is essentially the same.
Suppose to the contrary that $e$ does not divide $a$. Then when we try to divide $a$ by $e$, we get a positive remainder. More precisely,
$$a=qe+r,$$
where $0\lt r\lt e$. Then
$$r=a-qe=a-q(ax+by)=a(1-qx)+b(-qy).$$
This means that $r$ is a linear combination of $a$ and $b$, and is positive and less than $e$. This contradicts the fact that $e$ is the smallest positive linear combination of $a$ and $b$.
Set $c=\gcd(a,b)$ and $d=\gcd(a,b-a)$ for convenience.
Step 1: Prove that $c$ is a common divisor of $a,b-a$. This proves $c\le d$.
Step 2: Prove that $d$ is a common divisor of $a,b$. This proves $d\le c$.
Best Answer
Call the two numbers $x$ and $y$ and note that $\gcd(x,y)×\operatorname{lcm}(x,y)=xy$. Substituting what we know we find $$3×12x=xy$$ $$y=36=3×12$$ Then $x$ is the largest number less than 100, divisible by 3 and whose quotient after division by 3 is coprime with 12. 99 and 96 don't work because their quotients (33 and 32) are not coprime with 12, but 93 works. The largest possible sum is thus $93+36=129$.