Solving a linear Diophantine equation

divisibility. Euclidean algorithm:
\begin{align*}
1263 &= 3\cdot420+3,\\
420 &= 3\cdot140+0.
\end{align*}
So $\divisibility(1263,420)=3$, which divides $21$, so solutions exist.

Bézout. From the first line, $3=1263-3\cdot420$. Multiplying by $7$:
$$21=7\cdot1263-21\cdot420.$$
So $(x_0,y_0)=(7,-21)$ is a particular solution.

General solution. The homogeneous equation $1263x+420y=0$ gives $x=140m$, $y=-421m$ (after dividing by $\divisibility=3$). Hence
$$(x,y)=(7+140m,\,-21-421m),\qquad m\in\mathbb{Z}.$$

Remark. This is the corrected partiel exercise from Paris Diderot.