If the equation of the plane passing through the point$\left(1,1,2\right)$ and perpendicular to the line$\mathrm{x}-3\mathrm{y}+2\mathrm{z}-1=0=4\mathrm{x}-\mathrm{y}+\mathrm{z}$ is $\mathrm{Ax}+\mathrm{By}+\mathrm{Cz}=1$, then $140 (\mathrm{C}-\mathrm{B}+\mathrm{A})$is equal to______.

Dr’s line: 

${\overrightarrow{\mathrm{n}}}_1\times {\overrightarrow{\mathrm{n}}}_2=\left|\begin{array}{ccc}\widehat{\mathrm{i}}& \widehat{\mathrm{j}}& \widehat{\mathrm{k}}\\ 1& -3& 2\\ 4& -1& 1\end{array}\right|==-\widehat{\mathrm{i}}+7\widehat{\mathrm{j}}+11\widehat{\mathrm{k}}$

$\mathrm{drs}\text{ of normal to the plane is }-1,7,11$

$\mathrm{point} \mathrm{on} \mathrm{the} \mathrm{plane} \mathrm{is} (1,1,2)$

$\text{ Equation of plane : }-1(x-1)+7(y-1)+11(z-2)=0$

$-x+7y+11z=-1+7+22$

$-x+7y+11z=28$

$\frac{-\mathrm{x}}{28}+\frac{7\mathrm{y}}{28}+\frac{11}{28}\mathrm{z}=1$
$140 (C-B+A)=140(\frac{11}{28}-\frac{7}{28}-\frac{1}{28})=15$