In \(\mathrm{a}+\mathrm{b}+\mathrm{c}=0\), than the equation \(3 \mathrm{ax}^{2}+2 \mathrm{bx}+\mathrm{c}=0\) has, in the interval \((0,1)\) (a) at least one root (b) at most one root (c) no root (d) Exactly one root exist.

We have two equations: \[a + b + c = 0\] and \[3ax^2 + 2bx + c = 0\] Rewrite the quadratic equation to better study its properties: \[f(x) = 3ax^2 + 2bx + c\]

Compute the first derivative of f(x): \[f'(x) = \frac{d}{dx} (3ax^2 + 2bx + c)\] \[f'(x) = 6ax + 2b\] Critical points occur when the first derivative is equal to 0. Solve for x: \[6ax + 2b = 0\] \[x = -\frac{b}{3a}\]

Let's analyze the critical point -\(\frac{b}{3a}\) in respect to the interval (0,1). Since a, b, and c are not all zero, there are three possibilities: 1. a, b > 0 and c = -a - b < 0 2. a, b < 0 and c = -a - b > 0 3. a > 0 > b and c = -a - b For the first case, a, b > 0, we have the critical point -\(\frac{b}{3a}\) < 0 and f(1) = \(-a - a + 2b = b - a > 0\), which makes the curve positive outside of the interval (0,1). For this case, there's no root (`c`). For the second case, we have a and b < 0, the critical point -\(\frac{b}{3a}\) > 0 and f(1) = \(-a - a + 2b = b - a < 0\), which makes the curve negative outside of the interval (0,1). For this case, there's no root (`c`). For the third case, we have a > 0 > b, the critical point -\(\frac{b}{3a}\) lies inside the interval (0,1). Here, f(0) = c and f(1) = b - a != 0. Also, since -\(\frac{b}{3a}\) is in (0,1), we have exactly one root in the interval (`d`).

Based on our analysis in Step 3, options (a), (b), and (c) are incorrect, while option (d) is: (d) Exactly one root exists.