Because x-2 is the denominator x≠2. The expression under the square root must always be positive, so if within x+1, x-3, x-2 they must all be positive, or two must be negative and one positive.
If they are all positive, x+1≥0, x-3≥0, x-2, that is, x≥-1, x≥3, x>2. Since 3 is greater than both -1 and 2, it is sufficient to say x≥3, when all three will be positive (zero is counted as a positive number, applicable only in the numerator).
Now consider pairs (x+1, x-3), x+1<0⇒x<-1, x-3<0⇒x<3; x-2>0. But x cannot be greater than 2 and less than -1. Next pair: (x+1, x-2) x<-1 plus x-3≥0. Contradictory, so this combination isn't acceptable.
Last pair: (x-3, x-2), x<2, plus x≥-1, so -1≤x<2 is valid.
Therefore the domain is -1≤x<2 and x≥3, which is [-1,2) and [3,∞).