Try to pick solution from : xyz = 6= 1 * 2 * 3 , Now Possible combinations for (x, y, z) becomes (1, 2 , 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1), but if we try these combinations in x+y-z=4 and x²-y²+z²=-4, only one combinations satisfies it andi.e. ( 2, 3, 1). Hence Answer.

Given equations:
1. x + y - z = 4
2. x^2 - y^2 + z^2 = -4
3. xyz = 6

Solving the equations:

Equation 1: x + y - z = 4

Equation 2: x^2 - y^2 + z^2 = -4

Step 1: Simplify equation 2 by factoring the difference of squares:
(x + y)(x - y) + z^2 = -4

Step 2: Use equation 1 to substitute (x + y) with (4 + z):
(4 + z)(x - y) + z^2 = -4

Step 3: Expand the equation:
4x - 4y + zx - zy + z^2 = -4

Step 4: Rearrange the terms:
4x - 4y + zx - zy + z^2 + 4 = 0

Step 5: Simplify the equation by dividing all terms by 4:
x - y + (z/4)(x - y + z) + 1 = 0

Step 6: Rearrange the terms:
(x - y)(1 + (z/4)) + (x - y + z) = 0

Step 7: Factor out (x - y):
(x - y)(1 + (z/4) + 1) = 0

Step 8: Simplify the equation:
(x - y)(2 + (z/4)) = 0

Step 9: Set each factor equal to zero:
x - y = 0 --> x = y
2 + (z/4) = 0 --> z/4 = -2 --> z = -8

Step 10: Substitute the values of x and z into equation 1 to solve for y:
x + y - z = 4
y + y - (-8) = 4
2y + 8 = 4
2y = -4
y = -2

Conclusion: The values of x, y, and z are x = -2, y = -2, and z = -8.