The answer is not 4 !!

By the distance equation, $PQ = \sqrt{(0 - 12)^2 + (5 - 10)^2} = 13$ , $PR = \sqrt{(0 - t)^2 + (5 - 0)^2} = \sqrt{t^2 + 25}$ , and $QR = \sqrt{(12 - t)^2 + (10 - 0)^2} = \sqrt{t^2 - 24t + 244}$ .

By the law of cosines, $\cos \angle PRQ = \frac{PR^2 + QR^2 - PQ^2}{2 \cdot PR \cdot QR}$ , or $\cos \theta = \frac{(\sqrt{t^2 + 25})^2 + (\sqrt{t^2 - 24t + 244})^2 - 13^2}{2 \cdot \sqrt{t^2 + 25} \cdot \sqrt{t^2 - 24t + 244}}$ , which can be rearranged to $\cos^2 \theta = \frac{t^4 - 24t^3 + 244t^2 - 1200t + 2500}{t^4 - 24t^3 + 269t^2 - 600t + 6100}$ .

By implicit differentiation, $2 \cos \theta \sin \theta \frac{d\theta}{dt} = \frac{(t^4 - 24t^3 + 269t^2 - 600t + 6100)(4t^3 - 72t^2 + 488t - 1200) - (t^4 - 24t^3 + 244t^2 - 1200t + 2500)(4t^3 - 72t^2 + 538t - 600)}{(t^4 - 24t^3 + 269t^2 - 600t + 6100)^2}$ , and when $\frac{d\theta}{dt} = 0$ (for a maximum $\theta$ ), this simplifies to $(t + 12)(t^2 - 12t + 50)(t^2 + 24t - 194) = 0$ , which solves to $t = 13\sqrt{2} - 12 \approx \boxed{6.385}$ for $0 < t < 12$ .

For the point $R$ that maximizes $\angle PRQ$ , the circumcircle of triangle $PQR$ is tangent to the $x$ -axis.

Let $r$ be the radius of the circle. Then the center of the circle is $(t,r)$ , so $\begin{aligned} t^2 + (r - 5)^2 &= r^2, \\ (t - 12)^2 + (r - 10)^2 &= r^2. \end{aligned}$

We can solve for $t$ to get $-12 \pm 13 \sqrt{2}$ . The root that makes sense is $-12 + 13 \sqrt{2}$ .

Slope of $\overline {PR}=m_1=-\dfrac{5}{t}$ .

That of $\overline {QR}=m_2=\dfrac{10}{12-t}$

So $\tan \angle {PRQ}=\dfrac{m_1-m_2}{1+m_1m_2}=$

$\dfrac{5(t+12)}{t^2-12t+50}$

This will be extrimum when it's first differential coefficient with respect to $t$ is zero :

$\implies t^2+24t-194=0\implies t\approx 6.385, -30.685$

For the angle to be maximum , the second derivative should be negative. This happens for $t\approx \boxed {6.385}$ .