Step 1: Get the intersection by calculus-free elimination:
Instead of factoring, solve $\frac{\partial S}{\partial x} = 0$ and $\frac{\partial S}{\partial y} = 0$ for the pair of lines $S = 0$. The point where both vanish is the intersection.
$\frac{\partial S}{\partial x} = 2x - 6y + 10 = 0$, so $x - 3y + 5 = 0$.
$\frac{\partial S}{\partial y} = -6x + 10y - 14 = 0$, so $3x - 5y + 7 = 0$.
Step 2: Solve:
From the first, $x = 3y - 5$. Then $9y - 15 - 5y + 7 = 0$, so $4y = 8$, $y = 2$, $x = 1$.
Step 3: Write the axis-parallel pair:
The vertical line is $x = 1$ and the horizontal line is $y = 2$. Together: $(x-1)(y-2) = 0$, which expands to $xy - 2x - y + 2 = 0$.
Options (A), (B) and (C) place the point at $(-2,-1)$, $(-2,1)$ or $(2,-1)$ and so do not pass through $(1,2)$.
Final Answer:
$xy - 2x - y + 2 = 0$, option (D).
\[ \boxed{xy - 2x - y + 2 = 0 \text{ (D)}} \]