Your method matters

$3x^{2} + 6x + 3 + 3y^{2} - 6y + 3 + 4 +4x - 4y - 4xy = 0$

$= 3((1 +x)^{2}) + 3((1 - y)^{2}) + 4(1 + x)(1 - y) = 0$

$=(1 +x)^{2} + (1 - y)^{2} +2[(1+x)^{2} + (1 - y)^{2} + 2(1 + x)(1 - y)] = 0$

$( 1 + x)^{2} + (1 -y)^{2} + 2[(1+x) + (1 - y)]^{2} = 0$

$thus (x , y ) =( -1 , 1 )$

Since the determinant of the given equation is $0$ , it[the given equation] represents a pair of straight lines. Hence our next step should be to factorize it. Well, no! Our next step is to differentiate it !!

Differentiating with respect to $x$ [treating $y$ as a constant] $\cfrac{ d }{ dx }\left( 3{ x }^{ 2 }+3{ y }^{ 2 }-4xy+10x-10y+10 \right) \\ = 6x-4y+10\\ or\quad 3x-2y+5$

Differentiating with respect to $y$ [treating $x$ as a constant] $\cfrac{ d }{ dy }\left( 3{ x }^{ 2 }+3{ y }^{ 2 }-4xy+10x-10y+10 \right) \\ =\quad 6y-4x-10\\ or\quad 3y-2x-5$

And now don't tell me you can't solve a simultaneous equations in 2 unknown.

Multiplying $3x-2y+5$ by $2$ and $3y-2x-5$ by $3$ and then subtracting we get $5y=5$

Then we find $x$ which comes to be $-1$

hence $\boxed { 1-(-1)=2 }$

From the problem statement, rearrange the equation in the following form:

$3y^2-(4x+10)y+(3x^2+10x+10) = 0$ ,

and use the quadratic formula,

$y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ , with:

$a = 3$ , $b = -(4x+10)$ , $c = 3x^2+10x+10$ .

So, $y = \frac{(4x+10) \pm \sqrt{((4x+10))^2-4(3)(3x^2+10x+10)}}{6}$ , simplifying we get

$y = \frac{1}{3} (5+2x \pm \sqrt{5} \sqrt{-1-2 x-x^2})$ , further:

$y = \frac{1}{3}(2x+5) \pm \frac{\sqrt{5}}{3}(x+1)i$

The problem statement requires $(x,y) \in \mathbb{R}$ , hence we set $x=a=-1$ resulting in $y=b=1$ . Hence, $\boxed{b-a = 2}$ .