If $\overline{\mathrm{a}}=2\overline{\mathrm{i}}+\overline{\mathrm{j}}-3\overline{\mathrm{k}}, \overline{\mathrm{b}}=\overline{\mathrm{i}}-2\overline{\mathrm{j}}+\overline{\mathrm{k}}, \overline{\mathrm{c}}=-\overline{\mathrm{i}}+\overline{\mathrm{j}}-4\overline{\mathrm{k}}, \overline{\mathrm{d}}=\overline{\mathrm{i}}+\overline{\mathrm{j}}+\mathrm{p}\overline{\mathrm{k}},$ where p is a
positive real number and $\left|\left(\overline{\mathrm{a}}\times \overline{\mathrm{b}}\right)\times \left(\overline{\mathrm{c}}\times \overline{\mathrm{d}}\right)\right|=5\sqrt{114}$, then a value of p is

$\begin{array}{rcl}[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{d}}]& =& \left|\begin{array}{ccc}2& 1& -3\\ 1& -2& 1\\ 1& 1& \mathrm{P}\end{array}\right|=1(-5)-1\left(5\right)+\mathrm{P}(-5)\\ & =& -10-5\mathrm{P}\\ [\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}] & =& \left|\begin{array}{ccc}2& 1& -3\\ 1& -2& 1\\ -1& 1& -4\end{array}\right| =2\left(7\right)-1(-3) -3(-1)\\ & =& 14+3+3=20\end{array}$

$\begin{array}{rcl}(\overline{\mathrm{a}}\times \overline{\mathrm{b}})+(\overline{\mathrm{c}}\times \overline{\mathrm{d}})& =& [\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{d}}] \overline{\mathrm{c}}-[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}] \overline{\mathrm{d}}\\ & =& (-10-5\mathrm{P}) (-\mathrm{i}+\mathrm{j}-4\mathrm{k})-20(\mathrm{i}+\mathrm{j}+\mathrm{Pk})\\ & =& (-10+5\mathrm{P})\mathrm{i}+(-30-5\mathrm{P})\mathrm{j}+\left(40\right)\mathrm{k}\\ \left|(\overline{\mathrm{a}}\times \overline{\mathrm{b}})\times (\overline{\mathrm{c}}\times \overline{\mathrm{d}})\right|& =& 5\sqrt{114}\\ 25{(\mathrm{P}-2)}^2+25{(\mathrm{P}+6)}^2+25\left(64\right)& =& 25\left(114\right)\end{array}$

$\begin{array}{rcl}{\mathrm{P}}^2+4-4\mathrm{P}+{\mathrm{P}}^2+36+12\mathrm{P}+64& =& 114\\ 2{\mathrm{P}}^2+8\mathrm{P}-10& =& 0 \Rightarrow {\mathrm{P}}^2+4\mathrm{P}-5=0\\ (\mathrm{P}+5) (\mathrm{P}-1)& =& 0 \Rightarrow \mathrm{P}=1, -5\end{array}$