Vectors  $\overline{p}=\mathrm{a}\overline{\mathrm{i}}+\mathrm{b}\overline{\mathrm{j}}+\mathrm{c} \overline{\mathrm{k}}, \overline{\mathrm{q}}=\mathrm{d}\overline{\mathrm{i}}+3\overline{\mathrm{j}}+4\overline{k}$  and  $\overline{r}=3\overline{i}+\overline{j}-2\overline{k}$  forming a triangle ABC are such that $\overline{p}=\overline{q}+\overline{r}$. If the area of $\mathrm{△}\mathrm{ABC} \text{is }5\sqrt{6 }\mathrm{sq}\text{.}$. units, then the sum of the absolute values of a, b, c is

$\begin{array}{r}A=ai+bj+ck\\ B=di+3j+4k\\ C=3i+j-2k\end{array}$

$\begin{array}{r}\therefore ai+bj+ck=(d+3)i+4j+2k\\ \Rightarrow a=d+3,b=4,c=2\end{array}$

$B\times C=\left|\begin{array}{ccc}i& j& k\\ d& 3& 4\\ 3& 1& -2\end{array}\right|=-10i+(2d+12)j+(d-9)k$

Area of the $\mathrm{△}\mathrm{ABC}=\frac{1}{2}|\mathrm{B}\times \mathrm{C}|=\frac{1}{2}\left[\sqrt{100+(2\mathrm{d}+12{)}^2+(\mathrm{d}-9{)}^2}\right]=5\sqrt{6}$

$\begin{array}{r}\Rightarrow \sqrt{5d^2+30d+325}=10\sqrt{6}\Rightarrow 5d^2+30d+325=600\\ \Rightarrow 5d^2+30d-275=0\Rightarrow d^2+6d-55=0\\ \Rightarrow (d+11)(d-5)=0\Rightarrow d=5,-11\end{array}$

$d=-11,a=-8,b=4,c=2$