$$\begin{aligned} & \text { Given, } \vec{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \vec{b} \\ & \begin{aligned} & \Rightarrow \vec{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})-(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \vec{b}=0 \\ & \Rightarrow \vec{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})+\vec{b} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=0 \\ & \therefore(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}=-\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{A}}) \\ & \Rightarrow \quad (\vec{a}+\vec{b}) \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=0 \end{aligned} \end{aligned}$$

Hence, $$\vec{a}+\vec{b}$$ is collinear to $$2 \hat{i}+3 \hat{j}+4 \hat{k}$$

Let $$\vec{a}+\vec{b}=\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})\quad \text{... (i)}$$

$$\begin{array}{l} \Rightarrow & |\vec{a}+\vec{b}|=|\lambda| \sqrt{2^2+3^2+4^2} \\ \Rightarrow & \sqrt{29}=|\lambda| \sqrt{29} \quad \text { (Given, }|\vec{a}+\vec{b}|=\sqrt{29}) \\ \Rightarrow & |\lambda|=1 \\ \Rightarrow & |\lambda|= \pm 1 \\ \therefore & \vec{a}+\vec{b}= \pm(2 \hat{i}+3 \hat{j}+4 \hat{k}) \end{array}$$

$$\begin{aligned} \text { Now, } &(\vec{a}+\vec{b}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k}) \\ &= \pm(2 \hat{i}+3 \hat{j}+4 \hat{k}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k}) \\ &= \pm(-14+6+12) \\ &= \pm 4 \end{aligned}$$