When do the needles of clock meet (Where did I go wrong)? 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 ?
 A: First of all, I can tell you that the only times all of the three needles meet is at midnight and at noon!
To see this, notice that the hour and minute needle meet 11 times during a twelve hour period, with equal intervals, so you can calculate when those times are, and you will find that the second hand needle will never be at that very spot, unless it is at midnight or noon.
As far as your calculations go: there are 60 minutes in an hour so the speed of the minute needle should be 60 times that of the hour needle, and you have it 120 times as fast ... I think you used 24 hours to figure out the speed of the hour needle instead of 12 hours ...
Most importantly though, figuring out the LCM is not going to answer this question for you. For example, suppose we have three needles that move at respective speeds of 1,2, and 3 degrees per second. Ok, so their LCM is 6 ... Does this mean that they mean 6 seconds in? No! In one second the first will have moves 6 degrees, the second 12 degrees, and the third 18 degrees, so they are not at all at the same spot.  I actually have no idea what this LCM would even mean.
A: 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.
A: I think its totally depends on approximation. If you approximate .00833333..... to .008. 
You get answer as 6000 seconds. It means 60 minutes. (Near to original answer).
So try it approximating to .0083 and check.
