We are given points

$$ A(3,7,4),\quad B(5,-2,-3),\quad C(-4,5,6),\quad \text{and} \quad D(1,2,3). $$

The volume of the parallelepiped formed by the edges at point A, namely $$\vec{AB},\ \vec{AC},\ \text{and}\ \vec{AD},$$ is given by the absolute value of the scalar triple product:

$$ \text{Volume} = \left| \vec{AB} \cdot (\vec{AC} \times \vec{AD}) \right|. $$

Below are the steps to compute the volume:

Compute the vectors:

$$ \vec{AB} = B - A = (5-3,\ -2-7,\ -3-4) = (2,\ -9,\ -7). $$

$$ \vec{AC} = C - A = (-4-3,\ 5-7,\ 6-4) = (-7,\ -2,\ 2). $$

$$ \vec{AD} = D - A = (1-3,\ 2-7,\ 3-4) = (-2,\ -5,\ -1). $$

Find the cross product $$\vec{AC} \times \vec{AD}:$$

Using the determinant formula:

$$ \vec{AC} \times \vec{AD} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -7 & -2 & 2 \\ -2 & -5 & -1 \end{vmatrix}, $$

we compute each component:

$$ \text{i-component}: \quad (-2)(-1) - (2)(-5) = 2 + 10 = 12. $$

$$ \text{j-component}: \quad -\Big[ (-7)(-1) - (2)(-2) \Big] = -\Big[ 7 + 4 \Big] = -11. $$

$$ \text{k-component}: \quad (-7)(-5) - (-2)(-2) = 35 - 4 = 31. $$

So,

$$ \vec{AC} \times \vec{AD} = (12,\ -11,\ 31). $$

Calculate the scalar triple product:

Take the dot product of $$\vec{AB}$$ and $$\vec{AC} \times \vec{AD}$$:

$$ \vec{AB} \cdot (\vec{AC} \times \vec{AD}) = (2)(12) + (-9)(-11) + (-7)(31). $$

Compute each term:

$$2 \times 12 = 24,$$

$$-9 \times -11 = 99,$$

$$-7 \times 31 = -217.$$

Adding these:

$$ 24 + 99 - 217 = -94. $$

Take the absolute value to get the volume:

$$ \text{Volume} = | -94 | = 94 \quad \text{cubic units}. $$

Thus, the volume of the parallelepiped is $$\boxed{94}$$ cubic units, which corresponds to Option B.

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