Question

if x²+y²+z²=2(x-y-z)-3 find the value of 2x-3y+4z

Answer 1

x^2 + y^2 + z^2 = 2x - 2y - 2z - 1 - 1 - 1 => then:

(x^2 - 2x + 1) + (y^2 + 2y + 1) + (z^2 + 2z + 1) = 0

(x - 1)^2 + (y + 1)^2 + (z + 1)^2 = 0 => if the sums of squares is zero then each of squares is zero:

(x - 1)^2 = 0 => x = 1
(y + 1)^2 = 0 => y = -1
(z + 1)^2 = 0 => z = -1

hence:

2x - 3y + 4z = 2 + 3 - 4 = 1

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Answer 2

Here is your solution

=x^2 + y^2 + z^2 = 2(x - y - z) - 3
It can be written as:- 

x^2 + y^2 + z^2 = 2x - 2y - 2z - 1 - 1 - 1
=> then:- 

(x^2 - 2x + 1) +(y^2 + 2y + 1) + (z^2 + 2z + 1) = 0 

(x - 1)^2 + (y + 1)^2 + (z + 1)^2 = 0
=> if the sums of squares is zero then each of squares is zero: 

(x - 1)^2 = 0 => x = 1 
(y + 1)^2 = 0 => y = -1 
(z + 1)^2 = 0 => z = -1 

hence:-

2x - 3y + 4z = 2 + 3 - 4 = 1