If (x - 3)² + (y - 5)² + (z - 4)² = 0, then the value of $$\frac{{{x^2}}}{9}{\text{ + }}\frac{{{y^2}}}{{25}}{\text{ + }}\frac{{{z^2}}}{{16}}$$    is?

Solution (By Examveda Team)

$$\eqalign{ & {\left( {x - 3} \right)^2}{\text{ + }}{\left( {y - 5} \right)^2}{\text{ + }}{\left( {z - 4} \right)^2} = 0 \cr & \therefore {\left( {x - 3} \right)^2} = 0{\text{ }}x = 3 \cr & {\left( {y - 5} \right)^2} = 0{\text{ }}y = 5 \cr & {\left( {z - 4} \right)^2} = 0{\text{ }}z = 4{\text{ }} \cr & \therefore \frac{{{x^2}}}{9}{\text{ + }}\frac{{{y^2}}}{{25}}{\text{ + }}\frac{{{z^2}}}{{16}} \cr & \Rightarrow \frac{9}{9} + \frac{{25}}{{25}} + \frac{{16}}{{16}} \cr & \Rightarrow 3 \cr} $$