factors of 120 = 2*2*5*3
factors of 210 = 2*5*3*7
factors of 270 = 3*3*3*2*5
the factors common to all is 2,3,5
no of ways of selecting one out of the 3 common factors is 3C1 ( in other words, we can select 2 or 3 or 5)
no of ways of selecting two common factors out of 3 is 3C2 ( in other words we can select 2*3 or 2*5 or 3*5)
no of ways of selecting all three common factors is 3C3 ( in other words we can select 2*3*5)
all common factors greater than 1 is 3C1 or 3C2 or 3C3 = 3C1 + 3C2 + 3C3 =7
hence ans is D

May someone assesses me here if this a right way to solve ?

Yes it is another way to solve it from a different angle

Why isn't the answer 3 instead of 7?
2, 3, and 5 are the common factors so why is there a need to use a combination? Could someone explain it?

\(15\) is a common factor and we have to include that too. We need all common factors so all \(>1\) will be considered.

We see that all three numbers are multiples of 10. So 5 and 2 are common prime factors.

Now let us find the common prime factors of 12, 21 and 27.

12=2 x 2 x 3
21=7 x 3
27=3 x 3 x 3

Thus the prime factor common to the above three numbers is 3.

So, we the following prime factors common to 120, 210 and 270

2, 3 and 5

Now let us find the rest of the common factors by multiplying the common prime factors among themselves.

2 x 3 = 6
2 x 5 = 10
3 x 5 = 15

and finally,

2 x 3 x 5 = 30.

So the factors common to 120, 210 and 270 are 2, 3, 5, 6, 10, 15, 30

There are 7 of them

The answer is D