= Solution
Interpret the given focal point as the centre of <perspective projection>, $C=(0,6,-8)$. First translate it to the origin with $H(0,-6,8)$. The direction from $C$ to the world origin is then $(0,-6,8)$, of length 10. Choose the <rotation matrix> $G(\theta)$ so that this direction becomes $(0,0,10)$:
$$
\cos\theta=\frac45,\qquad \sin\theta=-\frac35,\qquad \theta=-\arctan\frac34.
$$
Indeed, the transformed $y$ component is $-6(4/5)-8(-3/5)=0$, and the transformed $z$ component is $(-6)(-3/5)+8(4/5)=10$. Thus the required mapping, followed by the <perspective divide>, is
$$
\boxed{M=K(5)\,G\!\left(-\arctan\frac34\right)\,H(0,-6,8).}
$$
For a world <point> $(x,y,z)$, its camera coordinates and projected device coordinates are
$$
X=x,\qquad Y=\frac45y+\frac35z,\qquad Z=10-\frac35y+\frac45z,
$$
$$
(u,v)=\left(\frac{5x}{10-3y/5+4z/5},\frac{4y+3z}{10-3y/5+4z/5}\right),\qquad Z>0.
$$
The world origin projects to $(0,0)$ and lies at camera depth 10. A <point> five units along the unit viewing direction from $C$ has camera coordinates $(0,0,5)$, so it lies on the central image plane as expected. Here device coordinates are image-plane coordinates centred on the optical axis; an actual <pixel> viewport would require its own specified scale and offset.
Only <rotations> about the $x$-axis are available. Consequently the camera's horizontal axis stays parallel to the world $x$-axis; the viewing direction lies in the world $yz$ plane. The available orientation has pitch but no independent yaw or roll. <Translations> move the camera and the projected origin but cannot supply the missing orientation degrees of freedom.
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