Question 886157
i found the solution on the web and i'll repeat it here.
x^2 + y^2 + z^2 = 2 * (x - y - z) - 3
simplify to get:
x^2 + y^2 + z^2 = 2x - 2y - 2z - 3
subtract 2x from both sides of the equation and add 2y and add 2z to both sides of the equation to get:
x^2 - 2x + y^2 + 2y + z^2 + 2z = -3
complete the squares on the left side of the equation to get:
(x-1)^2 + (y+1)^2 + (z+1)^2 = -3 + 1 + 1 + 1 which becomes:
(x-1)^2 + (y+1)^2 + (z+1)^2 = 0
if x = 1 and y = -1 and z = -1, then the sum of these terms will be equal to 0.
you will get:
0^2 + 0^2 + 0^2 = 0
so you have a solution of:
x = -1
y = 1
z = 1
the original equation of:
x^2 + y^2 + z^2 = 2x - 2y - 2z - 3
becomes:
(1)^2 + (-1)^2 + (-1)^2 = 2(1) - 2(-1) - 2(-1) - 3 which becomes:
1 + 1 + 1 = 2 + 2 + 2 - 3 which becomes:
3 = 6 - 3 which becomes:
3 = 3
when x = 1 and y = -1 and z = -1, the value of 2x - 3y + 4z becomes 2(1) - 3(-1) + 4(-1) which is equal to 2 + 3 - 4 which is equal to 5 - 4 which is equal to 1.