Can someone double check if my truth table make sense to the problem statement?
To solve
Create a Karnaugh Map (K-map) with your inputs and output.

Circle each group of 1's (or zeros) with the largest group you can until you have covered all 1's (or zeros).
Remember:

Use a Gray code for row/col headers (only one bit changes between neighbouring cells)
Circles can go over the end and wrap round.
Circles can overlap.
Make circles as large as you can
Circles can only contain power of 2 (1,2,4,8,etc…) numbers of cells


Write down what you see for each group, following these rules

If a variable changes in a group then it's not included in the term.
If a variables value is 0, then it's term is inverted
The terms are products (logical AND)
The result is the sum of products (logical OR)

e.g.

In the red group the C term is both 0 and 1 so C is not included.
In the green group, A changes so is not included.
In the blue group A and C are both 1 but B changes so it is not included.

RED:  A.B
GREEN: B.C
BLUE: A.C
The minimal result:
X = A.B + B.C + A.C  (1)
But you get
X = A'.B.C + A.B'.C + A.B.C' + A.B.C  (2)
Which, if you notice, is just the sum of the smallest possible circles.

Q: Why are (1) and (2) not the same?
A: They are the same!
(I will re-write this to make things easier to follow)
X = A.B.C' + A.B.C + A'.B.C + A.B'.C  (2)
To get from (1) to (2):
Remember A+A' = 1
So we can multiply out each term in (1) with the missing term and its negative, that's the same as ANDing with 1.
X = (A.B).(C+C') + (A+A').(B.C) + A.(B+B').C
X = A.B.C + A.B.C' + A.B.C + A'.B.C + A.B.C + A.B'.C
The A.B.C terms are common, so we can remove two of them and get
X = A.B.C' + A.B.C + A'.B.C + A.B'.C
Which is what we want.
Going the other way, from (2) to (1):
Our result is factors of ABC, so we need to remove all A'B'C' terms.
Group where ABC terms are common and add in more ABC terms if necessary
X = (A.B.C' + A.B.C) + (A'.B.C + A.B.C) + (A.B'.C + A.B.C)
X = (A.B).(C+C') + (A'+A).(B.C) + A.(B'+B).C
X = (A.B)(1) + (1)(B.C) + A.(1).C
X = A.B + B.C + A.C
Which is what we want.