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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 ?

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    I think this is not the only flaw (perhaps it is even irrelevant), but approximating a quatity before taking the $\operatorname{lcm}$ is a very bad idea. For instance, $\operatorname{lcm}(1000,1000000)=1000000$, but $\operatorname{lcm}(1000,1000001)=1000001000$ – 2017-01-31
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    Computing the LCM of these three different speeds is not going to tell you anything about where they meet as they go around ... What does this LCM even *mean*? – 2017-01-31
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    The LCM of the periods relative to the hour hand will, on the other hand, tell you when all three hands meet at once: The minute hand travels at one revolution per $12/11$ hours, relative to the hour hand; the second hand travels at one revolution per $6/359$ hours, relative to the hour hand. Since $359$ and $11$ are both primes, the LCM of these two fractions is simply $12$, meaning that the three hands can only meet once every $12$ hours. – 2017-01-31

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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.

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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.

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    Please help me in finding a way to mathematically deduce it, sir. – 2017-01-31
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    Furthermore, I have checked my calculations in calculator, I really don't know where i went wrong and it frustrates me. WHAT WENT WRONG !! – 2017-01-31
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    Oh, I see! I think you worked with 24 hours! It should be 12 hours. – 2017-01-31
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    Thanks, I will try that. (^o^) (Did you intend a pun there ?) – 2017-01-31
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    @ChandlerBing Unfortunately, that speed error was only a minor problem with your approach, as it is not at all clear how computing the LCM on the speeds of the needles is going to provide you with any answer to this question ... – 2017-01-31
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    If you take the common multiples of velocities, you get the time period at which they will meet, because they travel as a multiple of the speed in unit time. (I don't know how to explain it more clearly). It should work. :o idk why it ain't – 2017-02-01
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    @ChandlerBing OK, but did you see my example of having three needles move at 1,2, and 3 degrees per second? Since their LCM is 6, by your logic that would mean that 6 second into it, they should meet ... but obviously they don't! Rather than velocity, you need to take into account the 'distance' (i.e. amount of degrees) that they move, and the fact that they can go through 360 degree circles any number of times at which case they 'reset' back to 0 degrees. Indeed, if the needles didn't make multiple rounds, they would never meet. Your calculation doesn't take that into account at all. – 2017-02-01
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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.