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: 
**ANSWER IS NOT CORRECT**

You really just need a couple equations

```
h = hi + min/60

h/12 = min/60 = sec/60

```

hi is the integer hour before the time you want to find (hi=6)
h, min, sec are the actual hour, minute and second they intersect.

Substitute min/60 for h/12 and the equation becomes

```
h = hi + h/12 
(11/12)*h = hi 
h = 12/11*hi 
h = 12/11*6
**h = 72/11**

min = h/12*60
min = 6/11*60
sec = min = 6/11*60

h = 6.5454
min = 32.7272
sec = 32.7272

```

For an actual time drop the decimal points of the hour and minute and your left with

```
**final time is 6:32:32.727**

```

**EDIT TO SHOW THAT IT IS NOT CORRECT**

As per @Poolsharker answer these equations are incorrect another formula that should have been considered is

```
min = mini + sec/60
59/60*min = mini

```

From the previous final time mini = 32

```
min = 60/59*32
min = 32.5424 

```

which does not match the previous min time so they will not all 3 coincide
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).