Rotation in 3D

The axis of rotation must be perpendicular to both $\overrightarrow{AA'}$ and $\overrightarrow{BB'}$ , and hence must be parallel to the cross product of these vectors. Thus tge axis of rotation is parallel to the vector $\mathbf{j}$ . Since the perpendicular distances from $A$ and $A'$ to the axis of rotation must be the same, and similarly for $B$ and $B'$ , we deduce that the axis of rotation is the line $x=\alpha$ , $z=\beta$ , where $(1 - \alpha)^2 + \beta^2 \; = \; (2-\alpha)^2 + (1-\beta)^2 \hspace{2cm} (1-\alpha)^2 + (3-\beta)^2 \; = \; (5-\alpha)^2 + (1-\beta)^2$ and hence $\alpha=2$ , $\beta=0$ . Looking at the coordinates, it is now clear that the rotation is through $90^\circ$ , and hence the image of $C$ is the point $C'\;(8,5,-2)$ , making the answer $8 + 5 - 2 = \boxed{11}$ .