Given,

$$ \begin{aligned} & \vec{a}=\hat{i}+2 \hat{j}+3 \hat{k} \\\\ & \vec{b}=\hat{i}+\hat{j}-\hat{k} \\\\ & \vec{a} \cdot \vec{c}=11 \\\\ & \vec{b} \cdot(\vec{a} \times \vec{c})=27 \\\\ & \vec{b} \cdot \vec{c}=-\sqrt{3}|\vec{b}| \\\\ & (\vec{b} \times \vec{a}) \cdot \vec{c}=27 \end{aligned} $$

$$ \text { Let } \vec{c}=c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k} $$

As $$\vec{a} \cdot \vec{c}=11$$

$$ \therefore $$ $$ c_1+2 c_2+3 c_3=11 $$ ......(i)

Also, $\vec{b} \cdot \vec{c}=-\sqrt{3}|\vec{b}|$

$$ \begin{aligned} & \therefore c_1+c_2-c_3=-\sqrt{3} \sqrt{3} \\\\ & \Rightarrow c_1+c_2-c_3=-3 ......(ii) \end{aligned} $$

Also, $\vec{b} \cdot(\vec{a} \times \vec{c})=27$

$$ \therefore $$ $$ 5 c_1-4 c_2+c_3=27 $$ ...........(iii)

From (i), (ii) & (iii)

$$ \vec{c}=3 \hat{i}-2 \hat{j}+4 \hat{k} $$

$$ \begin{aligned} & |\vec{a} \times \vec{c}|^2=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 3 & -2 & +4 \end{array}\right|^2 \\\\ & =|14 \hat{i}+5 \hat{j}-8 \hat{k}|^2 \\\\ & =14^2+5^2+8^2=285 \end{aligned} $$