If (1,-1,2) and (2,1,-1) are the end points of a diameter of a...

The general equation $x^2+y^2+z^2+2 u x+2 v y+2 w z+d=0$ represents a sphere with centre $(-u,-v,-w) and radius$r=\sqrt{u^2+v^2+w^2-d}$If$B is the mid point of the line segment A C$where \mathrm{A}$=\left(x_1, y_1, z_1\right)$and$\mathrm{C}=\left(x_2, y_2, z_2\right) then. \mathrm{B}$=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)$ Given, equation of sphere $x^2+y^2+z^2+2 u x+ 2 v y+2 w z-1=0$ with AB as its diameter and $\mathrm{A}=(1,-1,2) and \mathrm{B}=(2,1,-1)$
$$\begin{aligned} \mathrm{C} & =\left(\frac{1+2}{2}, \frac{1-1}{2}, \frac{-1+2}{2}\right) \\ & =\left(\frac{3}{2}, 0, \frac{1}{2}\right) \end{aligned}$$ We know that centre of the sphere is given by $(-u,-v,-w)$ Comparing, then we get, $\left(\frac{3}{2}, 0, \frac{1}{2}\right)=(-u,-v,-w)$$\begin{aligned} u & =-\frac{3}{2}, v=0, w=-\frac{1}{2} \\ \therefore u+v+w & =-\frac{3}{2}+0-\frac{1}{2}=-\frac{4}{2}=-2\end{aligned}$$$