Step-by-step Solution:

\[ \text{Given, } \mathbf{a} \times (2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}) = (2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}) \times \mathbf{b} \] Using the property of cross products: \[ \mathbf{a} \times (2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}) = -\mathbf{b} \times (2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}) \] Rearranging, \[ (\mathbf{a} + \mathbf{b}) \times (2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}) = \mathbf{0} \] This implies that \( \mathbf{a} + \mathbf{b} \) is parallel to \( (2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}) \), so we write: \[ (\mathbf{a} + \mathbf{b}) = \lambda (2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}) \] Taking magnitudes on both sides, \[ |\mathbf{a} + \mathbf{b}| = |\lambda (2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k})| \] Since, \[ |(2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k})| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29} \] We get, \[ \sqrt{29} |\lambda| = \sqrt{29} \] \[ |\lambda| = 1 \] Thus, \[ \lambda = \pm 1 \] So, \[ (\mathbf{a} + \mathbf{b}) = \pm (2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}) \] Now, computing \( (\mathbf{a} + \mathbf{b}) \cdot (-7\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) \): \[ = \pm (2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}) \cdot (-7\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}) \] Using the dot product formula: \[ = \pm \left( (2 \times -7) + (3 \times 2) + (4 \times 3) \right) \] \[ = \pm \left( -14 + 6 + 12 \right) \] \[ = \pm 4 \] Thus, the final answer is: \[ \boxed{\pm 4} \]