The sum of the real roots of the equation $$\begin{vmatrix} x & -6 & -1 \\ 2 & -3x & x-3 \\ -3 & 2x & x+2 \end{vmatrix} = 0$$, is equal to:

We begin with the determinant

$$ \Delta \;=\; \begin{vmatrix} x & -6 & -1\\ 2 & -3x & x-3\\ -3 & 2x & x+2 \end{vmatrix}. $$

To obtain the characteristic polynomial, we expand this determinant along the first row. For a $$3\times3$$ matrix, the expansion formula is

$$ \begin{vmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{vmatrix} =\;a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}, $$

where each cofactor is $$C_{1j}=(-1)^{1+j}M_{1j}$$ and $$M_{1j}$$ is the minor obtained after deleting row 1 and column $$j$$.

We now compute each minor and cofactor step by step.

First cofactor $$C_{11}$$:

The minor is

$$ M_{11}=\begin{vmatrix} -3x & x-3\\ 2x & x+2 \end{vmatrix} =(-3x)(x+2)-(x-3)(2x) =-3x^2-6x-2x^2+6x =-5x^2. $$

Since $$(-1)^{1+1}=+1$$, we get $$C_{11}=M_{11}=-5x^2$$.

Thus the contribution from $$a_{11}=x$$ is

$$x\cdot C_{11}=x(-5x^2)=-5x^3.$$ Second cofactor $$C_{12}$$:

Delete column 2 to obtain

$$ M_{12}=\begin{vmatrix} 2 & x-3\\ -3 & x+2 \end{vmatrix} =2(x+2)-(x-3)(-3) =2x+4+3x-9 =5x-5 =5(x-1). $$

Because $$(-1)^{1+2}=-1$$, we have $$C_{12}=-M_{12}=-5(x-1)$$.

The entry in the first row and second column is $$-6$$, so the term generated is

$${(-6)}\cdot C_{12}=(-6)\bigl[-5(x-1)\bigr]=30(x-1).$$ Third cofactor $$C_{13}$$:

The needed minor is

$$ M_{13}=\begin{vmatrix} 2 & -3x\\ -3 & 2x \end{vmatrix} =2(2x)-(-3x)(-3) =4x-9x =-5x. $$

Here $$(-1)^{1+3}=+1$$, so $$C_{13}=M_{13}=-5x$$.

With $$a_{13}=-1$$, the contribution is

$${(-1)}\cdot C_{13}=(-1)(-5x)=5x.$$

Adding all three contributions, we get the determinant in simplified form:

$$ \Delta=-5x^3+30(x-1)+5x. $$

Open the parentheses and combine like terms:

$$ \Delta=-5x^3+30x-30+5x =-5x^3+35x-30. $$

Factor out $$-5$$:

$$ \Delta=-5\bigl(x^3-7x+6\bigr). $$

The determinant must vanish, so we set $$\Delta=0$$ and divide by $$-5$$ (which does not affect the roots):

$$ x^3-7x+6=0. $$

We now solve the cubic equation $$x^3-7x+6=0.$$ Possible rational roots are the factors of $$6$$, i.e. $$\pm1,\pm2,\pm3,\pm6$$.

Checking sequentially:

For $$x=1$$: $$1-7+6=0,$$ so $$x=1$$ is a root.

For $$x=2$$: $$8-14+6=0,$$ so $$x=2$$ is also a root.

For $$x=-3$$: $$-27+21+6=0,$$ hence $$x=-3$$ is a root as well.

Thus the cubic factors completely as

$$ x^3-7x+6=(x-1)(x-2)(x+3). $$

The real roots are therefore $$x=1,\;x=2,\;x=-3.$$

The sum of the real roots is

$$ 1+2+(-3)=0. $$

Hence, the correct answer is Option A.