Question: What will be the time when the hour, minute, and second hands will all coincide in between 6 O'Clock and 7 O'Clock?

How should I approach this type of question? Can anyone explain it?

Answer: 
From my calculations...

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> all three hands will not intersect 'exactly' between 6 O'clock and 7 O'clock
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The way I approached it isn't elegant but I think my calculations are correct:

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>  Clock face makes up 360 degrees. When the minute hand moves 360 degrees, the hour hand has moved 30 degrees (1/12). So I did my calculations by moving the minute hand to where the hour hand is, and then adjusting the hour hand because of the minute hand movement.  
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>  So at 6:00, hour hand (hh) is at 180 degrees, minute hand (mh) is at 0.  
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>  Move mh 180 degrees, move hh 180 \* (1/12) = 15 degrees.  
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>  Move mh 15 degrees, move hh 15 \* (1/12) = 1.25 degrees.  
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>  Move mh 1.25 degrees, move hh 1.25 (1/12) = 0.104 degrees. and so on...  
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>  The calculations start repeating at the hour hand having moved 16.3636. So if we are 16.3636 degrees between the 6 and the 7 and 30 degrees = 5 minutes, then 16.3636 / (30/5) = 2.7272 minutes. Now we convert 0.7272 into seconds (0.7272 \* 60) and we get 43.632. So I believe that the hour hand and minute hand we be nearly exact at 6:32:43.6, however that does leave the second hand outside. The best approximation would be 6:32:32.72 as mentioned by gtwebb.
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Side note

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>  While checking my calculations, I found this site calculating all the times the hour hand and minute hand intersect, good future reference <http://www.kodyaz.com/articles/how-many-times-a-day-clocks-hands-overlap.aspx>
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There's no three pointers coincidence between 6 o'clock and 7 o'clock.

We can solve this problem using equations of motion.

The equation of motion in terms of angles of a pointer is:

$$\theta=\theta\_i+\dot{\theta} t$$

where $\theta\_i$ is the initial angle (at $t=0$), $\dot{\theta}$ the angular frequency and, $t$ the time.

We have three bodies: hours ($h$), minutes ($m$) and seconds ($s$) pointers.

Consider that the initial condition is at 6 o'clock ($t=0$). For a polar referential system where $\theta=0$ in 12 o'clock, with a direct clock direction for this angle, we know that at the initial condiditon, we have:

$$ \theta\_h^i=\pi \text{ (rad)}, \hspace{5pt} \theta\_m^i=0 \text{ (rad)}, \hspace{5pt} \theta\_s^i=0 \text{ (rad)}$$ 

In relation to the angular frequencies, we have:

$$ \dot{\theta\_h}=\frac{2\pi}{12\times3600} \text{(rad/s)}, \hspace{5pt} \dot{\theta\_m}=\frac{2\pi}{3600} \text{ (rad/s)}, \hspace{5pt} \dot{\theta\_s}=\frac{2\pi}{60} \text{ (rad/s)}$$ 

The condition for three pointers coincidence is:

$$\theta\_h(t)=\theta\_h(t)-2\pi n=\theta\_h(t)- 2\pi m$$

where $n$ e $m$ are integers (the subtraction by $2\pi n$ and $2\pi m$ cancels the cyclicality of the pointers). We need that $n\in[0,1]$, $m\in[0,60]$ and $0\leq t\leq3600$ s for the coincidence be between 6 and 7 o'clock.

So, solving the system of equations of $\theta\_h(t)=\theta\_h(t)-2\pi n$ and $\theta\_h(t)-2\pi n=\theta\_h(t)- 2\pi m$ in order to $t$ am $m$ we get:

$$t=\frac{21600}{11}\left(1+2n\right) \text{ (s)},\hspace{15pt} m=\frac{719n+354}{11}$$

I tried to find the first value of $n$ that satisfies the conditions of $n$ and $m$ being positive integers, by substituition of $n=1,2,3...$ on the $m(n)$ expression and verified if $m$ was an integer. I got $n=5$ which means that $t=21600 s=6h$. Because $t=0$, at 6 o'clock, the coincidence happens at 12 o'clock, which means that it's impossible that it happens between 6 and 7 o'clock (remember that $t(n)$ is a crescent function with $n$, and $n=5$ is the lowest value for coincidence).