If the equation of the plane passing through the point ( 1, 1, 2 )...

Equation of the required plane is

$a\left(x - 1\right) + b\left(y - 1\right) + c\left(z - 2\right) = 0 . . . . \left(i\right)$

Since, required plane is perpendicular to the line $x - 3y + 2z - 1 = 0 = 4x - y + z$, therefore normal vector of the plane is

\overset{\to}{n} = \left|\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 1 & - 3 & 2 \\ 4 & - 1 & 1\end{cases}\right|

⇒ \overset{\to}{n} = - \overset{^}{i} + 7\overset{^}{j} + 11\overset{^}{k}

So, direction ratios of normal to the plane is $- 1, 7, 11$

Equation of plane is

$- 1 ( x - 1 ) + 7 ( y - 1 ) + 11 ( z - 2 ) = 0$

$⇒ - x + 7y + 11z = 28$

$⇒ \dfrac{x}{- 28} + \dfrac{7y}{28} + \dfrac{z}{\dfrac{28}{11}} = 1$

Comparing with $Ax + By + Cz = 1$, we get

$A = - \dfrac{1}{28}, B = \dfrac{7}{28}, C = \dfrac{11}{28}$

So,

$140\left(C - B + A\right) = 140\left(\dfrac{11}{28} - \dfrac{7}{28} - \dfrac{1}{28}\right)$

$⇒ 140\left(C - B + A\right) = 140 \times \dfrac{3}{28} = 15$