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

<p>Line of intersection of the planes $x - 3y + 2z - 1 = 0$ and $4x - y + z = 0$ is normal $(\overrightarrow n )$ to the required plane.</p> <p>$$\overrightarrow n = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & { - 3} & 2 \cr 4 & { - 1} & 1 \cr } } \right| = - \widehat i + 7\widehat j + 11\widehat k$$</p> <p>Equation of plane is</p> <p>$- x + 7y + 11z = \lambda$</p> <p>It passes through (1, 1, 2)</p> <p>$\therefore$ $\lambda = 28$</p> <p>So, the plane is</p> <p>$- x + 7y + 11z = 28$</p> <p>$\Rightarrow {{ - 1} \over {28}}x + {7 \over {28}}y + {{11} \over {28}}z = 1$</p> <p>$A = {{ - 1} \over {28}},B = {7 \over {28}},C = {{11} \over {28}}$</p> <p>$140(C - B + A) = 15$</p>