Question 118

If $$x^{2}+y^{2}+z^{2}=2(x-y-z)-3$$, then the value of $$2x-3y+4z$$ is [Assume that x, y, z are all real numbers) :

Solution

Expression : $$x^{2}+y^{2}+z^{2}=2(x-y-z)-3$$

=> $$x^2 - 2x + y^2 + 2y + z^2 + 2z + 3 = 0$$

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

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

=> $$x = 1 , y = -1 , z = -1$$

To find : $$2x-3y+4z$$

= $$(2*1) - (3* -1) + (4* -1)$$

= $$2+3-4 = 1$$

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