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

Find the plane through $$(1,1,2)$$ perpendicular to the line of intersection of the planes $$x - 3y + 2z - 1 = 0$$ and $$4x - y + z = 0$$.

To determine the direction of the line of intersection, note that the normals to the given planes are $$\vec{n_1} = (1, -3, 2)$$ for the first plane and $$\vec{n_2} = (4, -1, 1)$$ for the second plane. Hence the direction vector of the line is the cross product $$\vec{n_1} \times \vec{n_2}$$, which can be evaluated by the determinant

$$ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 4 & -1 & 1 \end{vmatrix} = \hat{i}(-3 - (-2)) - \hat{j}(1 - 8) + \hat{k}(-1 + 12) $$

and this yields the direction vector $$(-1, 7, 11)$$.

Since the required plane is perpendicular to this line, its normal vector is $$(-1, 7, 11)$$, and it passes through the point $$(1,1,2)$$. Therefore, its equation can be written as

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

which simplifies to

$$ -\,x + 7y + 11z = 28. $$

To express this in the form $$Ax + By + Cz = 1$$, divide both sides by 28, giving

$$ \frac{-1}{28}x + \frac{7}{28}y + \frac{11}{28}z = 1, $$

so that $$A = -\frac{1}{28}$$, $$B = \frac{1}{4}$$, and $$C = \frac{11}{28}$$. Finally, we compute

$$ C - B + A = \frac{11}{28} - \frac{1}{4} - \frac{1}{28} = \frac{11}{28} - \frac{7}{28} - \frac{1}{28} = \frac{3}{28}, $$

and hence

$$ 140 \times \frac{3}{28} = 15. $$

Therefore, the answer is $$15$$.