Problem 1191

1. We start with the given equation:
$$x^3 + y^3 + z^3 - 3xyz = 607$$
This can be factored using the identity for the sum of cubes:
$$x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx)$$
Therefore, we have:
$$(x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx) = 607$$

2. Since 607 is a prime number, the factors of 607 are 1 and 607. Thus, we have two cases to consider:
$$x+y+z = 607 \quad \text{and} \quad x^2 + y^2 + z^2 - xy - yz - zx = 1$$

3. Next, we multiply the second equation by 2:
$$2(x^2 + y^2 + z^2 - xy - yz - zx) = 2 \cdot 1 = 2$$
This simplifies to:
$$(x-y)^2 + (y-z)^2 + (z-x)^2 = 2$$

4. Since the sum of three squares is 2, the only possible values for the squares are 1, 1, and 0. Therefore, two of the variables must differ by 1, and the third must be equal to one of the other two. Without loss of generality, assume:
$$x = y \quad \text{and} \quad z = x + 1$$

5. Substitute $z = x + 1$ and $y = x$ into the first equation:
$$x + x + (x + 1) = 607$$
Simplify to find $x$:
$$3x + 1 = 607 \implies 3x = 606 \implies x = 202$$

6. Therefore, we have:
$$x = 202, \quad y = 202, \quad z = 203$$

7. To find the minimum value of $x + 2y + 3z$, substitute the values of $x$, $y$, and $z$:
$$x + 2y + 3z = 202 + 2(202) + 3(203)$$
Calculate step-by-step:
$$x + 2y + 3z = 202 + 404 + 609 = 1215$$

The final answer is $\boxed{1215}$