Given $\tan A+3\cos B+6\sin C=1;$ $3\tan A+\cos B+4\sin C=4$ and $5\tan A+3\cos B-8\sin C=-2.$

Let $\tan A=x, \cos B=y \& \sin C=z,$ then the given equations become

$x+3y+6z=1 ...\left(i\right)$

$3x+y+4z=4 ...\left(ii\right)$ and

$5x+3y-8z=-2 ...\left(iii\right)$

Let $P=\left[\begin{array}{ccc}1& 3& 6\\ 3& 1& 4\\ 5& 3& -8\end{array}\right], X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$ and $Q=\left[\begin{array}{c}1\\ 4\\ -2\end{array}\right].$

Then, the system of equations can be written as $PX=Q.$

Hence, we have $X=P^{-1}Q.$

Now $\left|P\right|=\left|\begin{array}{ccc}1& 3& 6\\ 3& 1& 4\\ 5& 3& -8\end{array}\right|$

$\Rightarrow \left|P\right|=1\left(-8-12\right)-3\left(-24-20\right)+6\left(9-5\right)$

$\Rightarrow \left|P\right|=-20+132+24=136.$

The cofactors of the elements of the matrix $P$ are

$A_{11}=-8-12=-20, A_{12}=-\left(-24-20\right)=44, A_{13}=9-5=4,$

$A_{21}=-\left(-24-18\right)=42, A_{22}=-8-30=-38, A_{23}=-\left(3-15\right)=12,$

$A_{31}=12-6=6, A_{32}=-\left(4-18\right)=14, A_{33}=1-9=-8.$Thus, the adjoint matrix of $P$ is

$adjP=\left[\begin{array}{ccc}-20& 42& 6\\ 44& -38& 14\\ 4& 12& -8\end{array}\right].$

And $P^{-1}=\frac{adjP}{\left|P\right|},$

$\Rightarrow P^{-1}=\frac{1}{136}\left[\begin{array}{ccc}-20& 42& 6\\ 44& -38& 14\\ 4& 12& -8\end{array}\right].$

The solution of the given system is

$\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{136}\left[\begin{array}{ccc}-20& 42& 6\\ 44& -38& 14\\ 4& 12& -8\end{array}\right]\left[\begin{array}{c}1\\ 4\\ -2\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{136}\left[\begin{array}{c}-20\times 1+42\times 4+6\times \left(-2\right)\\ 44\times 1+\left(-38\right)\times 4+14\times \left(-2\right)\\ 4\times 1+12\times 4+\left(-8\right)\times \left(-2\right)\end{array}\right]$

$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{1}{136}\left[\begin{array}{c}136\\ 136\\ 68\end{array}\right]$

$\Rightarrow x=\tan A=1, y=\cos B=-1$ and $z=\sin C=\frac{1}{2}.$

Also, $A$ and $C$ lie in $\left[0, \frac{\pi }{2}\right)$ and $B$ lies in $\left[0, 2\pi \right].$

Thus $\tan A=1, \Rightarrow A=\frac{\mathrm{\pi }}{4}, \cos B=-1, \Rightarrow B=\mathrm{\pi }$ and $\sin C=\frac{1}{2}, C=\frac{\mathrm{\pi }}{6}.$

Now $B-2 A-C=\mathrm{\pi }-2\times \frac{\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{6}=\frac{\mathrm{\pi }}{3}.$