We have,
$\mathbf{A B}=p \hat{\mathbf{i}}+q \hat{\mathbf{j}}+r \hat{\mathbf{k}}$
$\mathbf{A C}=s \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$
$\Rightarrow \quad \mathbf{C A}=-s \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$
and $\quad \mathbf{C B}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$
Area of $\triangle A B C$ is $5 \sqrt{6}$
$\Rightarrow \quad \frac{1}{2}|\mathbf{C A} \times \mathbf{C B}|=5 \sqrt{6}$ $\ldots(\mathrm{i})$
Now, $\mathbf{C A} \times \mathbf{C B}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ -s & -3 & -4 \\ 3 & 1 & -2\end{array}\right|$
$=\hat{\mathbf{i}}(6+4)-\hat{\mathbf{j}}(2 s+12)+\hat{\mathbf{k}}(-s+9)$
$=10 \hat{\mathbf{i}}-\hat{\mathbf{j}}(2 s+12)+\hat{\mathbf{k}}(-s+9)$
$\therefore \frac{1}{2} \sqrt{100+(2 s+12)^2+(-s+9)^2}=5 \sqrt{6}$
$\Rightarrow \sqrt{100+\left(4 s^2+144+48 s\right)+\left(s^2+81-18 s\right)}=10 \sqrt{6}$
$\Rightarrow \quad 325+5 s^2+30 s=600$
$\Rightarrow \quad 5 s^2+30 s-275=0$
$\Rightarrow \quad s^2+6 s-55=0$
$\Rightarrow \quad s^2+11 s-5 s-55=0$
$\Rightarrow \quad s(s+11)-5(s+11)=0$
$\Rightarrow \quad(s+11)(s-5)=0$
$\Rightarrow \quad s-5=0$ or $s+11 \neq 0$
$\Rightarrow \quad s=5$
Now, $\quad \mathbf{A B}=p \hat{\mathbf{i}}+q \hat{\mathbf{j}}+r \hat{\mathbf{k}}$
$\mathbf{B C}=-3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$
and $\quad \mathbf{C A}=-s \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$
By applying triangle law.
$\mathbf{A B}+\mathbf{B C}+\mathbf{C A}=\mathbf{O}$
$\Rightarrow(p \hat{\mathbf{i}}+q \hat{\mathbf{j}}+r \hat{\mathbf{k}})+(-3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})+(-5 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})$ $=\mathbf{O}$
$\Rightarrow(p-8) \hat{\mathbf{i}}+(q-4) \hat{\mathbf{j}}+(\mathbf{r}-2) \hat{\mathbf{k}}=\mathbf{O}$
$\Rightarrow \quad p=8, q=4$ and $r=2$