Let $\vec{a}=3 \hat{i}+2 \hat{j}+\hat{k}, \vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}$ and $\vec{c}$ be a vector such that $(\vec{a}+\vec{b}) \times \vec{c}=2(\vec{a} \times \vec{b})+24 \hat{j}-6 \hat{k}$ and $(\vec{a}-\vec{b}+\hat{i}) \cdot \vec{c}=-3$. Then $|\vec{c}|^2$ is equal to ________.

<p>$$\begin{aligned} & (\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=2(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})+24 \hat{\mathrm{j}}-6 \hat{\mathrm{k}} \\ & (5 \hat{\mathrm{i}}+\hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \times \overrightarrow{\mathrm{c}}=2(7 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-7 \hat{\mathrm{k}})+24 \hat{\mathrm{j}}-6 \hat{\mathrm{k}} \\ & \left|\begin{array}{lrr} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 5 & 1 & 4 \\ x & y & \mathrm{z} \end{array}\right|=14 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-20 \hat{\mathrm{k}} \\ & \Rightarrow \hat{\mathrm{i}}(\mathrm{z}-4 \mathrm{y})-\hat{\mathrm{j}}(5 \mathrm{z}-4 \mathrm{x})+\hat{\mathrm{k}}(5 \mathrm{y}-\mathrm{x})=14 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-20 \hat{\mathrm{k}} \\ & \mathrm{z}-4 \mathrm{y}=14,4 \mathrm{x}-5 \mathrm{z}=10,5 \mathrm{y}-\mathrm{x}=-20 \\ & (\mathrm{a}-\mathrm{b}+\mathrm{i}) \cdot \overrightarrow{\mathrm{c}}=-3 \\ & (2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \cdot \overrightarrow{\mathrm{c}}=-3 \\ & 2 x+3 y-2 z=-3 \\ & \therefore x=5, y=-3, z=2 \\ & |\overrightarrow{\mathrm{c}}|^2=25+9+4=38 \end{aligned}$$</p>