The sum of the real roots of the equation x - 6 - 1 2 - 3 x x - 3 - 3 2 x x + 2 = 0 , is equal to:

Given $\left|\begin{array}{ccc}x& -6& -1\\ 2& -3x& x-3\\ -3& 2x& x+2\end{array}\right|=0$

On expansion, we get $x\left\{-3x\left(x+2\right)-2x\left(x-3\right)\right\}+6\left\{2\left(x+2\right)+3\left(x-3\right)\right\}-1\left(4x-9x\right)=0$

$\Rightarrow x\left\{-3x^2-6x-2x^2+6x\right\}+6\left\{2x+4+3x-9\right\}-1\left(-5x\right)=0$

$\Rightarrow -5x^3+30x-30+5x=0$

$\Rightarrow 5x^3-35x+30=0$

$\Rightarrow x^3-7x+6=0$

$\Rightarrow \left(x-1\right)\left(x-2\right)\left(x+3\right)$

$\Rightarrow x=1$ or $x=2$ or $x=-3.$

Here all roots $\left\{1, 2, -3\right\}$ are real, hence the sum of the real roots is $1+2-3=0.$