Given vectors:

$$ \begin{aligned} \vec{A} &= 4\hat{i} - 5\hat{j} + 3\hat{k} \\ \vec{B} &= 2\hat{i} - 10\hat{j} - 7\hat{k} \\ \vec{C} &= 5\hat{i} + 7\hat{j} - 4\hat{k} \end{aligned} $$

(i) To find $(\vec{A} \times \vec{B}) \cdot \vec{C}$:

First, let's find the cross product $\vec{A} \times \vec{B}$:

$$ \begin{aligned} \vec{A} \times \vec{B} &= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -5 & 3 \\ 2 & -10 & -7 \end{vmatrix} \\ &= \hat{i} [(-5)(-7) - (3)(-10)] - \hat{j} [(4)(-7) - (3)(2)] + \hat{k} [(4)(-10) - (-5)(2)] \\ &= \hat{i} [35 + 30] - \hat{j} [-28 - 6] + \hat{k} [-40 + 10] \\ &= 65\hat{i} + 34\hat{j} - 30\hat{k} \end{aligned} $$

Now, calculate the dot product $(\vec{A} \times \vec{B}) \cdot \vec{C}$:

$$ \begin{aligned} (\vec{A} \times \vec{B}) \cdot \vec{C} &= (65\hat{i} + 34\hat{j} - 30\hat{k}) \cdot (5\hat{i} + 7\hat{j} - 4\hat{k}) \\ &= (65)(5) + (34)(7) + (-30)(-4) \\ &= 325 + 238 + 120 \\ &= 683 \end{aligned} $$

(ii) To find $\vec{A} \times (\vec{B} \times \vec{C})$:

First, let's find the cross product $\vec{B} \times \vec{C}$:

$$ \begin{aligned} \vec{B} \times \vec{C} &= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -10 & -7 \\ 5 & 7 & -4 \end{vmatrix} \\ &= \hat{i} [(-10)(-4) - (-7)(7)] - \hat{j} [(2)(-4) - (-7)(5)] + \hat{k} [(2)(7) - (-10)(5)] \\ &= \hat{i} [40 + 49] - \hat{j} [-8 + 35] + \hat{k} [14 + 50] \\ &= 89\hat{i} - 27\hat{j} + 64\hat{k} \end{aligned} $$

Now, calculate the cross product $\vec{A} \times (\vec{B} \times \vec{C})$:

$$ \begin{aligned} \vec{A} \times (\vec{B} \times \vec{C}) &= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & -5 & 3 \\ 89 & -27 & 64 \end{vmatrix} \\ &= \hat{i} [(-5)(64) - (3)(-27)] - \hat{j} [(4)(64) - (3)(89)] + \hat{k} [(4)(-27) - (-5)(89)] \\ &= \hat{i} [-320 + 81] - \hat{j} [256 - 267] + \hat{k} [-108 + 445] \\ &= -239\hat{i} + 11\hat{j} + 337\hat{k} \end{aligned} $$

Alternative Method using Vector Triple Product Identity:

$$ \vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C} $$

Let's calculate the dot products:

$$ \begin{aligned} \vec{A} \cdot \vec{C} &= (4)(5) + (-5)(7) + (3)(-4) = 20 - 35 - 12 = -27 \\ \vec{A} \cdot \vec{B} &= (4)(2) + (-5)(-10) + (3)(-7) = 8 + 50 - 21 = 37 \end{aligned} $$

Substituting these values:

$$ \begin{aligned} \vec{A} \times (\vec{B} \times \vec{C}) &= -27(2\hat{i} - 10\hat{j} - 7\hat{k}) - 37(5\hat{i} + 7\hat{j} - 4\hat{k}) \\ &= (-54\hat{i} + 270\hat{j} + 189\hat{k}) - (185\hat{i} + 259\hat{j} - 148\hat{k}) \\ &= (-54 - 185)\hat{i} + (270 - 259)\hat{j} + (189 + 148)\hat{k} \\ &= -239\hat{i} + 11\hat{j} + 337\hat{k} \end{aligned} $$