IMO it is E

stmt 1.
one of them is negative, this can lead to many situation

stmt 2. again either x or y is negative say x is negaive y is positive

x = -1 y = 5

z can have many values for z =-2 the question stem holds true as for z = 2

combining

either x or y is negative if x is negative z is positive so is y given |x-y| > |x -z| hence Z has to be in the middle. but same does not hold if y<0 x>0 and z>0

say y =-1 x = 5 z = 6

hence it is E

I used a very long method to solve this, but this is the only way I could think of this problem:

Based on the info given in the question stem, we can deduce 4 possibilities on the number line:

P1) Y____X___Z (Z on the right end, i.e not in between X and Y)
P2) Z___X____Y (Z on the left, i.e. not in between X and Y)
P3) X___Z____Y (Z in the middle, X on the left, Y on the right)
P4) Y____Z___X (Z in the middle, Y on the left, X on the right)

Evaluation each statement separately:

Statement 1: EITHER (X, Y and Z are all < 0) OR (X<0 or Y<0 or Z<0)

If X,Y and Z are all < 0, then P1, P2, P3 and P4 are possible.
If either X, Y or Z is <0, then P1, P2, P3 and P4 are possible.

Not Sufficient.

Statement 2: EITHER (X<0, Y>0) OR (X>0, Y<0)

If X<0, then P2 and P3 are possible
If Y<0, then P1 and P4 are possible.

Not Sufficient.

Evaluating both statements together:

Z must be positive, and either X<0 or Y<0. P1, P3 and P4 are possible. Both together are insufficient, answer E.