I tried finding it out mathematically by taking their velocities in degrees per second, and finding out the L.C.M of them which should give me the time period in which all the needles intersect.
Speed of seconds needle : 6°/sec
Speed of minutes needle : 0.1°/sec
Speed of hours needle : 0.008333...°/sec
I multiplied all of these by 100000 and got 600000, 10000, 833 (I rounded off 833 term)
I found the L.C.M and got 499800000
I divide the result by 10000 again and I get L.C.M of 6, 0.1, and 0.00833 as 4998
Which means the time period is 4998 seconds = 83.3 minutes, but that is not when the needles of clocks meet, they meet at around 66 minutes, so where did I go wrong ?
Bram28 is correct in saying that the three hands only meet at midnight and noon. You can extend the analysis below to confirm that.
Let's, therefore, just consider the hour and minute hands. The hour hand moves one revolution in $12$ hours, whereas a minute hand revolves once every hour, or $12$ times as fast. In other words, while the hour hand is making $x$ revolutions, the minute hand is making $12x$ revolutions. The minute hand "laps" the hour hand when $12x$ is also the same as $1+x$:
$$ 12x = 1+x $$
$$ 11x = 1 $$
or $x = 1/11$. Since the minute hand is making $12/x = 12/11$ revolutions, and each revolution is one hour, it takes $12/11$ hours for the two hands to meet again.
Some tedious but straightforward arithmetic shows that the second hand has made $720x/11 = 720/11$ revolutions in this time. Since $720$ divided by $11$ leaves a remainder of $5$, when the hour and minute hands first meet after $12$ o'clock, they are both $1/11$ of the way into their latest revolution, the second hand is $5/11$ of its way into its latest revolution. If you continue these computations, you will see that all three hands will not meet again until twelve hours have passed.