Question

if x, y and z are real numbers (x-3)² + (y-4)² + (z-5)² =0 then, find the value of (X+Y+Z) ?

Answer 1

$(x - 3) {}^{2} + (y - 4) {}^{2} + (z - 5) {}^{2} = 0$
since every term is square,, so each term will give a positive number..... and it is not possible that sum of 3 positive numbers is zero..... so the required condition is ::
$(x - 3) {}^{2} = 0 \\ \\ x = 3 \\ \\ (y - 4) {}^{2} = 0 \\ y = 4 \\ \\ (z - 5) {}^{2} = 0 \\ z = 5$

so,
$x + y + z = 3 + 4 + 5 = 12$


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