Let f(x) = x^3 + vmatrix x^2 & x & 1 f'(1) & f''(2) & f'''(3) 1 & 1...

Given $f(x) = x^3 + \begin{vmatrix} x^2 & x & 1 \\ f'(1) & f''(2) & f'''(3) \\ 1 & 1 & 0 \end{vmatrix}$

Expanding the determinant along the first row, we get:
$f(x) = x^3 + x^2(0 - f'''(3)) - x(0 - f'''(3)) + 1(f'(1) - f''(2))$
$\Rightarrow f(x) = x^3 - f'''(3)x^2 + f'''(3)x + f'(1) - f''(2)$

Differentiating with respect to $x$ successively:
$f'(x) = 3x^2 - 2f'''(3)x + f'''(3)$
$f''(x) = 6x - 2f'''(3)$
$f'''(x) = 6$

From this, we have $f'''(3) = 6$.

Substitute $f'''(3) = 6$ into the expression for $f''(x)$:
$f''(x) = 6x - 12$
At $x = 2$, $f''(2) = 12 - 12 = 0$.

Substitute $f'''(3) = 6$ into the expression for $f'(x)$:
$f'(x) = 3x^2 - 12x + 6$
At $x = 1$, $f'(1) = 3(1)^2 - 12(1) + 6 = -3$.

Now substitute the values of $f'(1)$, $f''(2)$, and $f'''(3)$ back into the expression for $f(x)$:
$f(x) = x^3 - 6x^2 + 6x + (-3) - 0$
$f(x) = x^3 - 6x^2 + 6x - 3$

To find $f(10)$:
$f(10) = (10)^3 - 6(10)^2 + 6(10) - 3$
$f(10) = 1000 - 600 + 60 - 3 = 457$.

Answer: $457$