3. Two numbers \(x\) and \(y\) satisfy the equation \(26 x^{2} + 23 x y - 3 y^{2} - 19 = 0\) and are the sixth and eleventh terms, respectively, of a decreasing arithmetic progression consisting of integers. Find the common difference of this progression.
Solution: Factorize the expression $26 x^{2}+23 x y-3 y^{2}$. For $y \neq 0$, we have $y^{2}\left(26\left(\frac{x}{y}\right)^{2}+23\left(\frac{x}{y}\right)-3\right)=26 y^{2}\left(\frac{x}{y}+1\right)\left(\frac{x}{y}-\frac{3}{26}\right)=(x+y)(26 x-3 y)$. This formula is valid for all real numbers $y$. According to the problem, the integers $x$ and $y$ satisfy the equation $(x+y)(26 x-3 y)=19$. The integers $x+y$ and $26 x-3 y$ are divisors of the number 19, and the following cases are possible:

1) $\left\{\begin{array}{c}x+y=19 \\ 26 x-3 y=1\end{array}\right.$
2) $\left\{\begin{array}{c}x+y=-19 \\ 26 x-3 y=-1\end{array}\right.$
3) $\left\{\begin{array}{c}x+y=-1 \\ 26 x-3 y=-19\end{array}\right.$
4) $\left\{\begin{array}{c}x+y=1 \\ 26 x-3 y=19\end{array}\right.$

We solve each system by the method of addition. Multiply both sides of the first equation of the system by 3 and add to the second:

1) $\left\{\begin{array}{l}29 x=58 \\ x+y=19\end{array}\right.$
2) $\left\{\begin{array}{c}29 x=-58 \\ x+y=-19\end{array}\right.$
3) $\left\{\begin{array}{l}29 x=-22 \\ x+y=-1\end{array}\right.$
4) $\left\{\begin{array}{l}29 x=22, \\ x+y=1 .\end{array}\right.$

Systems 3) and 4) do not have integer solutions. Since the progression is decreasing, it follows that $x<y$, which is satisfied by the solutions of system 2): $x=-2$, $y=-17$. Thus, $a_{6}=-2, a_{11}=-17$, or $a_{1}+5 d=-2, a_{1}+10 d=-17$. From this, we get $d=-3$.

Answer: $d=-3$.