Question: A,B and C start running on a circular track from the same point.
A and B run in a clockwise fashion.
C runs anti clockwise.
When A and C meet for the first time,
C is at a distance which is equal to quarter of circumference of circular track.
It is also known that A runs faster than C.
The ratio of speeds of A,C and B cannot be.

$\large \color{black}{ a.)\quad 5:1:2 \quad \hspace{.33em}\\~\\
b.)\quad 3:1:1 \quad\hspace{.33em}\\~\\
c.)\quad 4:2:1 \quad \hspace{.33em}\\~\\
d.)\quad 3:2:1 \quad \hspace{.33em}\\~\\
}$

options are showing ratio of speeds of $A:C:B$.

so far from the question i have drawn the two possible diagrams .
where B's speed can be $\text{
speed\_{B}}>\text{
speed\_{A}}$ or
$\text{
speed\_{A}}>\text{
speed\_{B}}$. 

and concluded that$ \text{
speed\_{A}}:\text{
speed\_{C}}::3:1$

![enter image description here](https://i.stack.imgur.com/dkzHw.png)

![enter image description here](https://i.stack.imgur.com/FrpP9.png)

screenshot of the question

![enter image description here](https://i.stack.imgur.com/M4Udf.jpg)

Answer: 
You are correct that the ratio of speeds of A and C is $3:1$ That shows that none of a,c,d are correct. b is still a possibility. Any of a,c,d should be an acceptable answer. I suspect it is a typo and instead of "cannot" it should be "might", when b would be the correct answer. My logic is just trying to find a simple rewording that yields a single correct answer.
I don’t think one can answer this MC question correctly. Assuming that this question is solvable, we can say that it has a unique solution (which should appear as one of the options) and this will leave the remaining three options become “cannot be’s”. This question is then equivalent to requiring to choose a “cannot be” from three “cannot be’s”. How?

---

However, I found this question has the following tricky part:-

“When A and C **meet** for the first time, C is **at a distance which is equal to quarter of circumference of circular track**.”

For simplicity, I let $O$ be the center of the circle whose radius is $10$ units, $P$ the starting point, $Q$ is at a quarter of the circle (counting from the anti-clockwise direction), and $R$ is at a distance of $5\pi$ from $P$.

![enter image description here](https://i.stack.imgur.com/FdNop.png)

Therefore, the meeting point is NOT $Q$ but $R$ where $PR =$ minor arc $(PQ) = \frac {2 \pi (10)}{4} = 5\pi$

The angles shown are calculate-able (but I let the machine do the job), such that $\angle POR = 103.44^0$ and reflex $\angle POR = 256.56^0$.

Thus, $\frac {Speed\_A}{Speed\_C} = \frac {256.56}{103.44} = \frac {5}{2}$ approximately

If I am right, the OP's suggestion must be wrong. Then, if I have to choose one option, I think (b) is the most unlikely "cannot be".

Also, $S$ could be a meeting point too but cannot satisfy the “Speed\_A > Speed\_C” requirement.