If $$\begin{vmatrix} x^2+x & x+1 & x-2 \\ 2x^2+3x-1 & 3x & 3x-3 \\ x^2+2x+3 & 2x-1 & 2x-1 \end{vmatrix} = ax - 12$$, then $$a$$ is equal to:

We are given the determinant equation:

$$\begin{vmatrix} x^2+x & x+1 & x-2 \\ 2x^2+3x-1 & 3x & 3x-3 \\ x^2+2x+3 & 2x-1 & 2x-1 \end{vmatrix} = ax - 12$$

We need to find the value of $$ a $$. To do this, we will compute the determinant of the given matrix and then compare it to the expression $$ ax - 12 $$.

The determinant $$ D $$ of a 3x3 matrix is calculated as follows:

$$D = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})$$

Here, $$ a_{11} = x^2 + x $$, $$ a_{12} = x + 1 $$, $$ a_{13} = x - 2 $$, $$ a_{21} = 2x^2 + 3x - 1 $$, $$ a_{22} = 3x $$, $$ a_{23} = 3x - 3 $$, $$ a_{31} = x^2 + 2x + 3 $$, $$ a_{32} = 2x - 1 $$, and $$ a_{33} = 2x - 1 $$.

First, compute $$ a_{22}a_{33} - a_{23}a_{32} $$:

$$a_{22}a_{33} - a_{23}a_{32} = (3x)(2x - 1) - (3x - 3)(2x - 1)$$

Factor out $$ (2x - 1) $$:

$$= (2x - 1)[3x - (3x - 3)] = (2x - 1)(3x - 3x + 3) = (2x - 1)(3) = 3(2x - 1)$$

Next, compute $$ a_{21}a_{33} - a_{23}a_{31} $$:

$$a_{21}a_{33} - a_{23}a_{31} = (2x^2 + 3x - 1)(2x - 1) - (3x - 3)(x^2 + 2x + 3)$$

Expand $$ (2x^2 + 3x - 1)(2x - 1) $$:

$$= 2x^2(2x - 1) + 3x(2x - 1) - 1(2x - 1) = (4x^3 - 2x^2) + (6x^2 - 3x) + (-2x + 1) = 4x^3 - 2x^2 + 6x^2 - 3x - 2x + 1 = 4x^3 + 4x^2 - 5x + 1$$

Expand $$ (3x - 3)(x^2 + 2x + 3) $$:

$$= 3(x - 1)(x^2 + 2x + 3) = 3[x(x^2 + 2x + 3) - 1(x^2 + 2x + 3)] = 3[x^3 + 2x^2 + 3x - x^2 - 2x - 3] = 3[x^3 + x^2 + x - 3] = 3x^3 + 3x^2 + 3x - 9$$

Subtract:

$$(4x^3 + 4x^2 - 5x + 1) - (3x^3 + 3x^2 + 3x - 9) = 4x^3 + 4x^2 - 5x + 1 - 3x^3 - 3x^2 - 3x + 9 = (4x^3 - 3x^3) + (4x^2 - 3x^2) + (-5x - 3x) + (1 + 9) = x^3 + x^2 - 8x + 10$$

Now, compute $$ a_{21}a_{32} - a_{22}a_{31} $$:

$$a_{21}a_{32} - a_{22}a_{31} = (2x^2 + 3x - 1)(2x - 1) - (3x)(x^2 + 2x + 3)$$

We already have $$ (2x^2 + 3x - 1)(2x - 1) = 4x^3 + 4x^2 - 5x + 1 $$.

Expand $$ (3x)(x^2 + 2x + 3) $$:

$$= 3x(x^2 + 2x + 3) = 3x^3 + 6x^2 + 9x$$

Subtract:

$$(4x^3 + 4x^2 - 5x + 1) - (3x^3 + 6x^2 + 9x) = 4x^3 + 4x^2 - 5x + 1 - 3x^3 - 6x^2 - 9x = (4x^3 - 3x^3) + (4x^2 - 6x^2) + (-5x - 9x) + 1 = x^3 - 2x^2 - 14x + 1$$

Now, substitute into the determinant formula:

$$D = (x^2 + x) \cdot [3(2x - 1)] - (x + 1) \cdot [x^3 + x^2 - 8x + 10] + (x - 2) \cdot [x^3 - 2x^2 - 14x + 1]$$

Expand each term:

First term: $$ (x^2 + x) \cdot 3(2x - 1) = 3(x^2 + x)(2x - 1) $$

Expand $$ (x^2 + x)(2x - 1) $$:

$$= x^2(2x - 1) + x(2x - 1) = 2x^3 - x^2 + 2x^2 - x = 2x^3 + x^2 - x$$

So, $$ 3(2x^3 + x^2 - x) = 6x^3 + 3x^2 - 3x $$

Second term: $$ - (x + 1)(x^3 + x^2 - 8x + 10) $$

Expand $$ (x + 1)(x^3 + x^2 - 8x + 10) $$:

$$= x(x^3 + x^2 - 8x + 10) + 1(x^3 + x^2 - 8x + 10) = x^4 + x^3 - 8x^2 + 10x + x^3 + x^2 - 8x + 10 = x^4 + 2x^3 - 7x^2 + 2x + 10$$

So, $$ - (x^4 + 2x^3 - 7x^2 + 2x + 10) = -x^4 - 2x^3 + 7x^2 - 2x - 10 $$

Third term: $$ (x - 2)(x^3 - 2x^2 - 14x + 1) $$

Expand $$ (x - 2)(x^3 - 2x^2 - 14x + 1) $$:

$$= x(x^3 - 2x^2 - 14x + 1) - 2(x^3 - 2x^2 - 14x + 1) = x^4 - 2x^3 - 14x^2 + x - 2x^3 + 4x^2 + 28x - 2 = x^4 - 4x^3 - 10x^2 + 29x - 2$$

Now, add all three terms together:

$$D = (6x^3 + 3x^2 - 3x) + (-x^4 - 2x^3 + 7x^2 - 2x - 10) + (x^4 - 4x^3 - 10x^2 + 29x - 2)$$

Combine like terms:

• $$ x^4 $$ terms: $$ -x^4 + x^4 = 0 $$
• $$ x^3 $$ terms: $$ 6x^3 - 2x^3 - 4x^3 = (6 - 2 - 4)x^3 = 0x^3 $$
• $$ x^2 $$ terms: $$ 3x^2 + 7x^2 - 10x^2 = (3 + 7 - 10)x^2 = 0x^2 $$
• $$ x $$ terms: $$ -3x - 2x + 29x = (-3 - 2 + 29)x = 24x $$
• Constant terms: $$ -10 - 2 = -12 $$

So, $$ D = 24x - 12 $$.

The equation is given as $$ D = ax - 12 $$. Comparing:

$$24x - 12 = ax - 12$$

Therefore, $$ a = 24 $$.

Hence, the correct answer is Option B.