If $\;$ $A \left(-1, 4, -3\right)$ $\;$ is one end of diameter $AB$ of the sphere $\;$ $x^2 + y^2 + z^2 -3x -2y + 2z - 15 = 0$ $\;$ then find the coordinates of $B$.
Equation of given sphere: $\;$ $x^2 + y^2 + z^2 - 3x -2y + 2z - 15 = 0$ $\;\;\; \cdots \; (1)$
Comparing equation $(1)$ with the general equation of sphere
$x^2 + y^2 + z^2 + 2ux + 2 vy + 2 wz + d = 0$ $\;$ gives
$u = \dfrac{-3}{2}, \;\; v = -1, \;\; w = 1, \;\; d = -15$
Center of the sphere $= \left(-u, -v, -w\right) = \left(\dfrac{3}{2}, 1, -1\right)$
Let the coordinates of point $B$ be $\left(\alpha, \beta, \gamma\right)$
$\because$ $\;$ $AB$ is the diameter of the sphere, we have
$\dfrac{-1 + \alpha}{2} = \dfrac{3}{2}$ $\implies$ $\alpha = 4$
$\dfrac{4 + \beta}{2} = 1$ $\implies$ $\beta = -2$
$\dfrac{-3 + \gamma}{2} = -1$ $\implies$ $\gamma = 1$
$\therefore$ $\;$ The coordinates of point $B$ are $\left(4, -2, 1\right)$