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: 
When we are a fraction $f$ of the way through a 12-hour "day" -- so $f=1$ means 12 hours have passed since noon or midnight -- the positions of the hands as fractions of a whole turn are: $f$ (for the hour hand), $12f$ (for the minute hand), and $720f$ (for the second hand), where integer differences are not visible. So for all three to coincide we need the differences $11f$ and $719f$ to be integers. Since 11 and 719 are coprime, this requires $f$ to be an integer, which means that the only 3-way coincidences happen with all hands pointing to the 12.

In particular, there is none between 6 and 7 o'clock.

(The bit about 11 and 719 being coprime may not be perfectly clear. Suppose $11f=m$ and $719f=n$ where $m,n$ are integers. Then $11n=719m$, so $11|719m$; since 11 is prime this requires either $11|719$ (no!) or $11|m$. So write $m=11k$; then $11f=m=11k$ so $f=k$ and $f$ is an integer.)
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).