Wee hours

Kind of an infuriating problem. Start with a clock at exactly $3:00$ . Let $m$ be the minutes from start, and ${M}_{1}, {M}_{2}, {M}_{3}$ be the positions of the hour, minute, and second hand respectively. Then

${ M }_{ 1 }=15+\dfrac { m }{ 12 }$
${ M }_{ 2 }=m$
${ M }_{ 3 }=60\left(m-n\right)$

where $n$ is the integer part of the minutes elapsed. We solve for $m, n$ the equation, for the first instance where both sides $>0$

${ M }_{ 2 }-{ M }_{ 1 }={ M }_{ 3 }-{ M }_{ 2 }$

with the solution $m=17.3027...$ and $n=17$ , and the number of seconds is $60m=1038.16...$

The positions of the hands of the clock can be represented as $h$ , $m/5$ , and $s/5$ , respectively, where one revolution corresponds to 12 units.

We know that $m = 60(h-H)\ \ \ \ s = 60(m-M)$ where $H = \lfloor h\rfloor$ and $M = \lfloor m\rfloor$ are positive integers.

The problem describes the situation in which $s/5 - m/5 = m/5 - h \ \ \ (= d > 0).$ Substituting the expression for $s$ , we get $12(m-M) - m/5 = m/5 - h\ \ \ \therefore\ \ \ 11\tfrac35 m + h = 12M;$ substituting for $m$ this becomes $11\tfrac35\cdot 60(h-H) + h = 12 M\ \ \ \therefore\ \ \ 697h = 12 M + 696 H.$ Thus $h = \frac{12M + 696 H}{697},\ \ \ m/5 = \frac{144M - 12H}{697},\ \ \ s/5 = \frac{276M - 720 H}{697},$ and it easy to check that the angle between the hands is $d = s/5 - m/5 = m/5 - h = \frac{132M - 708 H}{697}.$

Since it is shortly after 3 o'clock, we have $H = 3$ . For $M$ we need the smallest value that makes $d$ positive. The inequality becomes $d > 0 \ \ \ \therefore\ \ \ 132 M > 708 H \ \ \ \therefore M > 5\tfrac4{11}H,$ in this case, $M > 16\tfrac1{11}$ . Therefore the smallest solution after 3 o'clock is found when $M = 17$ . We calculate $h = \frac{12\cdot 17 + 696\cdot 3}{697} = 3.28838;$ subtracting 3 and multiplying by 3600 seconds per hour we find that this is $0.28838\cdot 3600 = \boxed{1038}$ seconds after 3 o'clock.

It was a nice problem... Really hats off to you man