Below is a proof that works in any GCD domain, using the universal definitions of GCD, LCM. These ideas go back to Euclid, who defined the greatest common measure of line segments. Nowadays this can be viewed more generally in terms of fractional ideals or Krull's $v$-ideals.
Theorem $\rm\ \ \ \left(\dfrac{a}b,\,\dfrac{A}B\right)\: =\: \dfrac{(a,A)}{[b,B]}\ \ $ if $\rm\ \ (a,b) = 1 = (A,B),\ \ $ where $\rm\ \ \begin{eqnarray} (a,b) &:=&\rm\ gcd\rm(a,b)\\\ \rm [a,b]\, &:=&\rm\ lcm(a,b)\end{eqnarray}$
Proof
$\rm\begin{eqnarray} &\rm\quad c &|&\rm a/b,\,A/B \\
\quad\iff&\rm Bbc &|&\rm\ aB,\,Ab \\
\iff&\rm Bbc &|&\rm (aB,\,\color{#C00}A\color{#0A0}b) \\
\iff&\rm Bbc &|&\rm (aB,\, (\color{#C00}A,aB)\,(\color{#0A0}b,aB))\ \ &\rm by\quad (x,\color{#C00}y\color{#0A0}z) = (x,\,(\color{#C00}y,x)\,(\color{#0A0}z,x)),\ \ see\ [1] \\
\iff&\rm Bbc &|&\rm (aB,\, (A,a)\, (b,B))\ \ &\rm by\quad (a,b) = 1 = (A,B) \\
\iff&\rm Bbc &|&\rm (a,A)\,(b,B)\ \ &\rm by\quad (A,a)\ |\ a,\ (b,B)\ |\ B \\
\iff&\rm\quad c &|&\rm (a,A)/[b,B]\ \ &\rm by\quad (b,B)\:[b,B] = bB, \ \ see\ [2]
\end{eqnarray}$
Here are said links to proofs of the gcd laws used: law [1] and law [2].
Let the highest exponent of prime $p$ that divides $a_1,b_1,c_1$ respectively be
$$A_1,B_1,C_1$$
WLOG $A_1\ge B_1\ge C_1$
So, the highest exponent of $p$ divides $a_2,b_2,c_2,a_3,b_3,c_3$ will respectively be $$C_1,C_1,B_1,B_2,A_2,A_2$$
The highest exponent of $p$ in gcd$(b_3,c_3)=A_2$ which is the same in $c_3$
Now this will hold true for any prime that divides $a_1b_1c_1$
Best Answer
You rewrite it as $(2021 - d)ab = 2021$. Now use the factorization of $2021$ given in the hint.