If 9x^2+4y^2+z^2=12x-12y+8z-29 , then find the value of xy+z .

The correct answer is Option (4) → 3

Moving every term to one side lets the expression be written as a sum of three perfect squares equal to zero. A square is never negative, so each square must be zero, and that fixes \(x\), \(y\) and \(z\).

• \(9x^2-12x+4y^2+12y+z^2-8z+29=0\)

• Splitting 29 as \(4+9+16\): \((9x^2-12x+4)+(4y^2+12y+9)+(z^2-8z+16)=0\)

• \((3x-2)^2+(2y+3)^2+(z-4)^2=0\)

• \(3x-2=0\Rightarrow x=\displaystyle\frac{2}{3}\); \(2y+3=0\Rightarrow y=-\displaystyle\frac{3}{2}\); \(z-4=0\Rightarrow z=4\)

• \(xy=\displaystyle\frac{2}{3}\times\left(-\frac{3}{2}\right)=-1\), so \(xy+z=-1+4=3\)