Let $\vec{a}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{d}}=\vec{a} \times \overrightarrow{\mathrm{b}}$. If $\overrightarrow{\mathrm{c}}$ is a vector such that $\vec{a} \cdot \overrightarrow{\mathrm{c}}=|\overrightarrow{\mathrm{c}}|$, $|\overrightarrow{\mathrm{c}}-2 \vec{a}|^2=8$ and the angle between $\overrightarrow{\mathrm{d}}$ and $\overrightarrow{\mathrm{c}}$ is $\frac{\pi}{4}$, then $|10-3 \overrightarrow{\mathrm{~b}} \cdot \overrightarrow{\mathrm{c}}|+|\overrightarrow{\mathrm{d}} \times \overrightarrow{\mathrm{c}}|^2$ is equal to _________.

<p>$$\begin{aligned} & \overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{~b}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{~d}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} \\ & =-\hat{\mathrm{i}}+\hat{\mathrm{j}} \\ & |\overrightarrow{\mathrm{c}}-2 \overrightarrow{\mathrm{a}}|^2=8 \\ & |\mathrm{c}|^2+4|\mathrm{a}|^2-4(\mathrm{a} \cdot \mathrm{c})=8 \\ & \mathrm{c}^2+12-4 \mathrm{c}=8 \\ & \mathrm{c}^2-4 \mathrm{c}+4=0 \\ & |\mathrm{c}|=2 \\ & \overrightarrow{\mathrm{~d}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} \\ & \overrightarrow{\mathrm{~d}} \times \overrightarrow{\mathrm{c}}=(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})^2 \times \overrightarrow{\mathrm{c}} \\ & \left(|\mathrm{~d}||\mathrm{c}| \sin \frac{\pi}{4}\right)^2=((\mathrm{a} \cdot \mathrm{c}) \cdot \mathrm{b}-(\mathrm{b} \cdot \mathrm{c}) \cdot \mathrm{a})^2 \\ & 4=4 \mathrm{~b}^2+(\mathrm{b} \cdot \mathrm{c})^2 2(\mathrm{a})^2-2(\mathrm{~b} \cdot \mathrm{c})(\mathrm{a} \cdot \mathrm{~b}) \end{aligned}$$</p> <p>$$\begin{aligned} &\text { Let } \mathrm{b} . \mathrm{c}=\mathrm{x}\\ &\begin{aligned} & 4=36+3 x^2-20 x \\ & 3 x^2-20 x+32=0 \\ & 3 x^2-12 x-8 x+32=0 \\ & x=\frac{8}{3}, 4 \\ & \text { b.c }=\frac{8}{3}, 4 \\ & \text { b.c }=\frac{8}{3} \\ & \text { Now }|10-3 \mathrm{~b} . \mathrm{c}|+|\mathrm{d} \times \mathrm{c}|^2 \\ & |10-8|+(2)^2 \\ & \Rightarrow 6 \text { Ans. } \end{aligned} \end{aligned}$$</p>