HINT:
First check if the greatest common divisor of 121 and 561 is a multiple of 13200. If so then this equation has a solution, otherwise it doesn't. Also it would be wise to divide by the greatest common divisor, so you'll work with smaller numbers.
And to obtain solutions apply the Euclidean Division Algorith. So you have:
$$561 = 4 \cdot 121 + 77$$
$$121 = 1 \cdot 77 + \cdots$$
Can you spot the pattern? Then going backwards you should be able to obtain one solution and quite easily a closed form of the solution. Then you can check when both solutions are positive.
Also I'm pretty sure there are a lot of nice books and video tutorials about solving a linear Diophantene equation on the internet. This one is quite simple and easy to follow.
Equation if we write in the General form:
$$aX^2+bXY+cY^2=eZ^2+jZW+tW^2$$
If in this equation there any equivalent to a quadratic form in which the root is an integer.
$$q=\sqrt{b^2+4a(e+j+t-c)}$$
Then there are solutions. They can be written by making the replacement.
$$x=(b(2(e+j+t)-b)+4ac)s-(b+2a)(j+2t)k$$
$$y=(b^2+4c(e+j+t-a))s^2-2(b+2c)(j+2t)sk+(j^2+4t(a+b+c-e))k^2$$
Then decisions can be recorded and they are as follows:
$$X=(b-2(e+j+t-c)\pm{q})p^2+2(q((j+2t)k-(b+2c)s)\pm{x})pn+$$
$$+(((2(e+j+t-c)-b)\pm{q})y+2((j+2t)k-(b+2c)s)x)n^2$$
$$***$$
$$Y=(\pm{q}-(b+2a))p^2+2(q((j+2t)k-(2(e+j+t-a)-b)s)\pm{x})pn+$$
$$+(((b+2a)\pm{q})y+2((j+2t)k-(2(e+j+t-a)-b)s)x)n^2$$
$$***$$
$$Z=(\pm{q}-(b+2a))p^2+2(q((j+2t)k-(b+2c)s)\pm{x})pn+$$
$$+(((b+2a)\pm{q})y+2((j+2t)k-(b+2c)s)x)n^2$$
$$***$$
$$W=(\pm{q}-(b+2a))p^2+2(q((2(a+b+c-e)-j)k-(b+2c)s)\pm{x})pn+$$
$$+(((b+2a)\pm{q})y+2((2(a+b+c-e)-j)k-(b+2c)s)x)n^2$$
$p,n,k,s $ - integers asked us.
Best Answer
The equation gives: $14y = xy - 14x$. So solving for $x$ and get: $x = 14 + \frac{196}{y-4}$.
So $x$ is an integer if $y - 4$ divides $196$. We can take it from here....