Step-by-step Solution:

1. Analyze the given vector equation. We are given the equation: $\overrightarrow a \times \left( {2\widehat i + 3\widehat j + 4\widehat k} \right) = \left( {2\widehat i + 3\widehat j + 4\widehat k} \right) \times \overrightarrow b$ Let's define a vector $\overrightarrow c = 2\widehat i + 3\widehat j + 4\widehat k$. The equation can be written as: $\overrightarrow a \times \overrightarrow c = \overrightarrow c \times \overrightarrow b$

2. Simplify the vector equation. Using the anti-commutative property of the cross product, which states that $\overrightarrow x \times \overrightarrow y = - (\overrightarrow y \times \overrightarrow x)$, we can rewrite the right side of the equation: $\overrightarrow c \times \overrightarrow b = - (\overrightarrow b \times \overrightarrow c)$ Substituting this back into our equation, we get: $\overrightarrow a \times \overrightarrow c = - (\overrightarrow b \times \overrightarrow c)$ Rearranging the terms to one side: $\overrightarrow a \times \overrightarrow c + \overrightarrow b \times \overrightarrow c = \overrightarrow 0$ Using the distributive property of the cross product, we can factor out $\overrightarrow c$: $(\overrightarrow a + \overrightarrow b) \times \overrightarrow c = \overrightarrow 0$

3. Interpret the result of the simplified equation. The cross product of two non-zero vectors is the zero vector ($\overrightarrow 0$) if and only if the two vectors are parallel. Let $\overrightarrow r = \overrightarrow a + \overrightarrow b$. Then our equation is $\overrightarrow r \times \overrightarrow c = \overrightarrow 0$. This implies that the vector $(\overrightarrow a + \overrightarrow b)$ is parallel to the vector $\overrightarrow c = 2\widehat i + 3\widehat j + 4\widehat k$. Therefore, we can write $(\overrightarrow a + \overrightarrow b)$ as a scalar multiple of $\overrightarrow c$: $\overrightarrow a + \overrightarrow b = \lambda \overrightarrow c = \lambda (2\widehat i + 3\widehat j + 4\widehat k)$ for some scalar $\lambda$.

4. Use the given magnitude to find the scalar $\lambda$. We are given that $|\overrightarrow a + \overrightarrow b| = \sqrt{29}$. From our previous step, we have $|\overrightarrow a + \overrightarrow b| = |\lambda \overrightarrow c| = |\lambda| |\overrightarrow c|$. Let's calculate the magnitude of $\overrightarrow c$: $|\overrightarrow c| = |2\widehat i + 3\widehat j + 4\widehat k| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$ Now, substitute the magnitudes back into the equation: $\sqrt{29} = |\lambda| \sqrt{29}$ This implies that $|\lambda| = 1$, so $\lambda = 1$ or $\lambda = -1$.

5. Determine the possible expressions for $(\overrightarrow a + \overrightarrow b)$. Based on the two possible values of $\lambda$, we have two possible expressions for the vector sum $(\overrightarrow a + \overrightarrow b)$:

   • Case 1: If $\lambda = 1$, then $\overrightarrow a + \overrightarrow b = 1 \cdot (2\widehat i + 3\widehat j + 4\widehat k) = 2\widehat i + 3\widehat j + 4\widehat k$.
   • Case 2: If $\lambda = -1$, then $\overrightarrow a + \overrightarrow b = -1 \cdot (2\widehat i + 3\widehat j + 4\widehat k) = -2\widehat i - 3\widehat j - 4\widehat k$.

6. Calculate the required dot product. We need to find a possible value of $(\overrightarrow a + \overrightarrow b) \cdot (-7\widehat i + 2\widehat j + 3\widehat k)$. Let's calculate this for both cases.

   • For Case 1: $(\overrightarrow a + \overrightarrow b) \cdot (-7\widehat i + 2\widehat j + 3\widehat k) = (2\widehat i + 3\widehat j + 4\widehat k) \cdot (-7\widehat i + 2\widehat j + 3\widehat k)$ $= (2)(-7) + (3)(2) + (4)(3)$ $= -14 + 6 + 12$ $= 4$

   • For Case 2: $(\overrightarrow a + \overrightarrow b) \cdot (-7\widehat i + 2\widehat j + 3\widehat k) = (-2\widehat i - 3\widehat j - 4\widehat k) \cdot (-7\widehat i + 2\widehat j + 3\widehat k)$ $= (-2)(-7) + (-3)(2) + (-4)(3)$ $= 14 - 6 - 12$ $= -4$

7. Conclusion. The possible values for the dot product are $4$ and $-4$. The question asks for a possible value. Looking at the options: A: $0$ B: $3$ C: $4$ D: $8$

   The value $4$ is listed as option C. Thus, a possible value is $4$.