Maximum $z$ -coordinate

Let $f(x,y,z)= (3y-2z)^2 + (z-3x)^2 + (2x-y)^2$ and $S=\{(x, y,z)| \; f\; \text{reaches its maximum value at}\; (x, y, z)\; \text{on the unit sphere (center at (0, 0, 0) and radius 1)}\} .$ Let $\vec{A}=<x, y, z>,$ where $x^2+y^2+z^2=1,$ and $\vec{B}=< 1, 2, 3 >.$ Then $f(x, y, z)=|\vec{A} \times \vec{B}|^2.$ Since, $| \vec{A} \times \vec{B} |=|\vec{A}| | \vec{B}| \sin \theta,$ where the angle $\theta$ is the angle formed by the vectors $\vec{A}$ and $\vec{B},$ then the $f(x, y, z) =14 \sin^2 \theta.$ Therefore, the maximum of $f(x, y, z)$ on the unit sphere is 14 and it is attained at $\theta=\frac{\pi}{2},$ that is, at any point $(x, y, z)$ belonging to the sphere $x^2+y^2+z^2=1$ such that $\vec{A} \cdot \vec{B}=x+2y+3z=0.$ Therefore, the set $S$ is a circle of radius 1, with center at $(0,0,0)$ in the plane $x+2y+3z =0.$ Hence the point with the highest $z$ -coordinate in $S$ satisfies the property that it is in the plane that contains $(1,2,3)$ and the $z$ -axis. The equation of this plane is $2x-y=0.$ Since this point is in both planes $x+2y+3z=0$ and $2x-y=0,$ then it is on the line passing through $(0, 0, 0),$ whose direction is given by the vector $< 2, -1, 0> \times <1, 2, 3>= < -3, -6, 5>$ . Using that the distance from this point to $(0,0,0)$ is 1, then the point has to be $(\frac{-3}{\sqrt{70}}, \frac{-6}{\sqrt{70}}, \frac{5}{\sqrt{70}}).$ So the maximum value of $z$ is $\frac{5}{\sqrt{70}}=\sqrt\frac{5}{14}.$ We can conclude that $A=5,$ $B=14$ and the answer for the problem is $\boxed{19}$ .

Remark: Instead of using the vector approach you can also use the Lagrange Identity to determine the set $S.$