Question
If (x - 5)2 + (y - 2)2 + (z - 9)2 = 0, then the value of (x + y - z) is?
Answer: Option C
$$\eqalign{
& {\left( {x - 5} \right)^2} + {\left( {y - 2} \right)^2} + {\left( {z - 9} \right)^2} = 0 \cr
& {\text{It is possible only when }} \cr
& x - 5 = 0 \cr
& x = 5 \cr
& y - 2 = 0 \cr
& y = 2 \cr
& z - 9 = 0 \cr
& z = 9 \cr
& \therefore x + y - z \cr
& = 5 + 2 - 9 \cr
& = 7 - 9 \cr
& = - 2 \cr} $$
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$$\eqalign{
& {\left( {x - 5} \right)^2} + {\left( {y - 2} \right)^2} + {\left( {z - 9} \right)^2} = 0 \cr
& {\text{It is possible only when }} \cr
& x - 5 = 0 \cr
& x = 5 \cr
& y - 2 = 0 \cr
& y = 2 \cr
& z - 9 = 0 \cr
& z = 9 \cr
& \therefore x + y - z \cr
& = 5 + 2 - 9 \cr
& = 7 - 9 \cr
& = - 2 \cr} $$
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