Question 1209900
Let the given equation be
$$x^2 + 3xy + 2y^2 = 1 - 12x + 5y$$
We can factor the left side as
$$(x+y)(x+2y) = 1 - 12x + 5y$$
Let $u = x+y$ and $v = x+2y$. Then $u+y=x+y$ so $y = v-u$ and $x = u-y = u-(v-u) = 2u-v$.
Substituting into the given equation, we have
$$uv = 1 - 12(2u-v) + 5(v-u)$$
$$uv = 1 - 24u + 12v + 5v - 5u$$
$$uv = 1 - 29u + 17v$$
$$uv - 17v = 1 - 29u$$
$$v(u-17) = 1 - 29u$$
If $u \neq 17$, we can solve for $v$:
$$v = \frac{1-29u}{u-17}$$
Then $y = v-u = \frac{1-29u}{u-17} - u = \frac{1-29u-u(u-17)}{u-17} = \frac{1-29u-u^2+17u}{u-17} = \frac{-u^2-12u+1}{u-17}$
And $x = 2u-v = 2u - \frac{1-29u}{u-17} = \frac{2u(u-17)-(1-29u)}{u-17} = \frac{2u^2-34u-1+29u}{u-17} = \frac{2u^2-5u-1}{u-17}$

Since $x$ and $y$ are real numbers, we must have $u \neq 17$.
We are interested in the possible values of $u = x+y$.
The equation can be rewritten as
$$x^2 + 3xy + 2y^2 + 12x - 5y - 1 = 0$$
Consider this as a quadratic in $x$:
$$x^2 + (3y+12)x + (2y^2-5y-1) = 0$$
For $x$ to be real, the discriminant must be non-negative:
$$(3y+12)^2 - 4(2y^2-5y-1) \ge 0$$
$$9y^2 + 72y + 144 - 8y^2 + 20y + 4 \ge 0$$
$$y^2 + 92y + 148 \ge 0$$
The roots of $y^2 + 92y + 148 = 0$ are
$$y = \frac{-92 \pm \sqrt{92^2 - 4(148)}}{2} = \frac{-92 \pm \sqrt{8464 - 592}}{2} = \frac{-92 \pm \sqrt{7872}}{2} = -46 \pm \sqrt{1968}$$
$$y = -46 \pm 4\sqrt{123}$$
Thus $y \le -46 - 4\sqrt{123}$ or $y \ge -46 + 4\sqrt{123}$.
Consider the equation as a quadratic in $y$:
$$2y^2 + (3x-5)y + (x^2+12x-1) = 0$$
The discriminant must be non-negative:
$$(3x-5)^2 - 4(2)(x^2+12x-1) \ge 0$$
$$9x^2 - 30x + 25 - 8x^2 - 96x + 8 \ge 0$$
$$x^2 - 126x + 33 \ge 0$$
The roots of $x^2 - 126x + 33 = 0$ are
$$x = \frac{126 \pm \sqrt{126^2 - 4(33)}}{2} = \frac{126 \pm \sqrt{15876 - 132}}{2} = \frac{126 \pm \sqrt{15744}}{2} = 63 \pm \sqrt{3936}$$
Thus $x \le 63 - \sqrt{3936}$ or $x \ge 63 + \sqrt{3936}$.

If $u=17$, then $0=1-29(17)=-492$, which is impossible. So $u\ne 17$ for all real solutions.
We need to find the range of $u$.
Let $y=ax+b$.
Then $x^2+3x(ax+b)+2(ax+b)^2=1-12x+5(ax+b)$.
$x^2+3ax^2+3bx+2a^2x^2+4abx+2b^2=1-12x+5ax+5b$.
$(1+3a+2a^2)x^2+(3b+4ab+12-5a)x+2b^2-5b-1=0$.
For real solutions, the discriminant must be non-negative.
$(3b+4ab+12-5a)^2-4(1+3a+2a^2)(2b^2-5b-1)\ge 0$.
$u=x+y=x+ax+b=(1+a)x+b$.
We want to find $u$.
If $x+y=17$, then $v=\frac{1-29(17)}{17-17}$. This is impossible.

We have $u=x+y$. Let $x+y=c$ and $x=c-y$.
$(c-y)^2+3(c-y)y+2y^2=1-12(c-y)+5y$.
$c^2-2cy+y^2+3cy-3y^2+2y^2=1-12c+12y+5y$.
$c^2+cy=1-12c+17y$.
$y(c-17)=c^2+12c-1$.
$y=\frac{c^2+12c-1}{c-17}$.
$x=c-y = c-\frac{c^2+12c-1}{c-17} = \frac{c(c-17)-c^2-12c+1}{c-17} = \frac{-29c+1}{c-17}$.
For real $x$ and $y$, $c\ne 17$.
Therefore $x+y$ can be any real number except 17.

Final Answer: The final answer is $\boxed{(-\infty,17)\cup(17,\infty)}$