Let $$\overrightarrow{\mathrm{a}}=4 \hat{i}-\hat{j}+\hat{k}, \overrightarrow{\mathrm{b}}=11 \hat{i}-\hat{j}+\hat{k}$$ and $\overrightarrow{\mathrm{c}}$ be a vector such that $$(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{c}} \times(-2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}})$$. If $(2 \vec{a}+3 \vec{b}) \cdot \vec{c}=1670$, then $|\vec{c}|^2$ is equal to:

<p>$$\begin{aligned} & \left.\begin{array}{l} \vec{a}=4 \hat{i}-\hat{j}+\hat{k} \\ \vec{b}=11 \hat{i}-\hat{j}+\hat{k} \end{array}\right] \\ & \vec{a} \cdot \vec{b}=44+1+1=46 \\ & |\vec{a}|^2=18,\left|\vec{b}^2\right|=123 \\ & (\vec{a}+\vec{b}) \times \vec{c}=\vec{c} \times(-2 \vec{a}+3 \vec{b}) \\ & (2 \vec{a}+3 \vec{b}) \cdot \vec{c}=1670 \\ & (\vec{a}+\vec{b}) \times c=(2 \vec{a}-3 \vec{b}) \times \vec{c} \\ & (-\vec{a}+4 \vec{b}) \times \vec{c}=0 \\ & \vec{c}=\lambda(4 \vec{b}-\vec{a}) \\ & (2 \vec{a}+3 \vec{b}) \cdot \lambda(4 \vec{b}-\vec{a})=1670 \\ & \lambda\left(5 \vec{a} \cdot \vec{b}-2|\vec{a}|^2+12|\vec{b}|^2\right)=1670 \\ & \lambda=\frac{1670}{5 \times 46-2 \times 18+12 \times 123} \\ & \lambda=1 \\ & \vec{c}=4 \vec{b}-\vec{a} \\ & =4(11 \hat{i}-\hat{j}+\hat{k})-(4 \hat{i}-\hat{j}+\hat{k}) \\ & =40 \hat{i}-3 \hat{j}+3 \hat{k} \\ & \left|\vec{c}^2\right|=1600+9+9 \\ & =1618 \\ \end{aligned}$$</p>