Let $\mathrm{ABC}$ be a triangle of area $15 \sqrt{2}$ and the vectors $$\overrightarrow{\mathrm{AB}}=\hat{i}+2 \hat{j}-7 \hat{k}, \overrightarrow{\mathrm{BC}}=\mathrm{a} \hat{i}+\mathrm{b} \hat{j}+\mathrm{c} \hat{k}$$ and $$\overrightarrow{\mathrm{AC}}=6 \hat{i}+\mathrm{d} \hat{j}-2 \hat{k}, \mathrm{~d}>0$$. Then the square of the length of the largest side of the triangle $\mathrm{ABC}$ is _________.

<p>Area of triangle $A B C=15 \sqrt{2}$</p> <p>$$\begin{aligned} & \Rightarrow \frac{1}{2}|\overline{A B} \times \overline{A C}|=15 \sqrt{2} \quad \text{.... (i)}\\ & \quad \overline{A B} \times \overline{A C}\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -7 \\ 6 & d & -2 \end{array}\right| \\ & =(7 d-4) \hat{i}-40 \hat{j}+(d-12) \hat{k} \quad \text{... (ii)} \end{aligned}$$</p> <p>From (i) and (ii) $5 d^2-8 d-4=0$</p> <p>$\Rightarrow d=\frac{-}{5}$ (Rejected) or $d=2$</p> <p>Also, $\overline{A B}+\overline{B C}=\overline{A C}$</p> <p>$$\begin{aligned} & \Rightarrow a+1=6 \Rightarrow a=5 \\ & b+2=d \Rightarrow b=0 \end{aligned}$$</p> <p>and $c-7=-2 \Rightarrow c=5$</p> <p>$|\overline{A B}|=\sqrt{54},|\overline{A C}|=\sqrt{44},|\overline{B C}|=\sqrt{50}$</p> <p>Largest side has length of $\sqrt{54}$ units</p>