1. Let $D(x)=\begin{vmatrix} x & -6 & -1 \\ 2 & -3x & x-3 \\ -3 & 2x & x+2 \end{vmatrix}.$ We need the real roots of $D(x)=0$, then their sum.

2. Expand the determinant along the first row: $D(x)=x\begin{vmatrix} -3x & x-3 \\ 2x & x+2 \end{vmatrix}-(-6)\begin{vmatrix} 2 & x-3 \\ -3 & x+2 \end{vmatrix}+(-1)\begin{vmatrix} 2 & -3x \\ -3 & 2x \end{vmatrix}.$

So, $D(x)=x\left[(-3x)(x+2)-2x(x-3)\right]+6\left[2(x+2)-(-3)(x-3)\right]-\left[2(2x)-(-3)(-3x)\right].$

3. Simplify each bracket:

For the first: $(-3x)(x+2)-2x(x-3)=-3x^2-6x-2x^2+6x=-5x^2.$ Hence, $x(-5x^2)=-5x^3.$

For the second: $2(x+2)-(-3)(x-3)=2x+4+3x-9=5x-5,$ so $6(5x-5)=30x-30.$

For the third: $2(2x)-(-3)(-3x)=4x-9x=-5x.$ Thus, $-\left(-5x\right)=+5x.$

4. Therefore, $D(x)=-5x^3+30x-30+5x=-5x^3+35x-30.$ Factor out $-5$: $D(x)=-5(x^3-7x+6).$ So the equation becomes $x^3-7x+6=0.$

5. Factor the cubic. Try rational roots $\pm 1, \pm 2, \pm 3, \pm 6$.

At $x=1$: $1-7+6=0,$ so $(x-1)$ is a factor.

Divide: $x^3-7x+6=(x-1)(x^2+x-6).$ Now, $x^2+x-6=(x+3)(x-2).$ Hence, $x^3-7x+6=(x-1)(x+3)(x-2).$

6. Thus the real roots are