Find the equation of the sphere passing through \( P(-2,3,4) \) and \( \mathrm{Q}(6,-1,3) \) with its center at the midpoint of PQ . The standard equation of the sphere is (Simplify your answer.)

To find the equation of the sphere, we first determine the midpoint \( M \) of points \( P(-2, 3, 4) \) and \( Q(6, -1, 3) \). The midpoint \( M \) is calculated as: \[ M = \left( \frac{-2 + 6}{2}, \frac{3 + (-1)}{2}, \frac{4 + 3}{2} \right) = \left( \frac{4}{2}, \frac{2}{2}, \frac{7}{2} \right) = (2, 1, \frac{7}{2}). \] Next, we determine the radius \( r \) of the sphere, which is the distance from the center \( M \) to either point \( P \) or \( Q \). We'll find the distance from \( M \) to \( P \): \[ r = \sqrt{(2 - (-2))^2 + (1 - 3)^2 + \left(\frac{7}{2} - 4\right)^2} = \sqrt{(2 + 2)^2 + (1 - 3)^2 + \left(\frac{7}{2} - \frac{8}{2}\right)^2}. \] Calculating each term: \[ = \sqrt{(4)^2 + (-2)^2 + \left(-\frac{1}{2}\right)^2} = \sqrt{16 + 4 + \frac{1}{4}} = \sqrt{20 + \frac{1}{4}} = \sqrt{\frac{80}{4} + \frac{1}{4}} = \sqrt{\frac{81}{4}}. \] This simplifies to: \[ r = \frac{9}{2}. \] Now, we can write the standard equation of the sphere with center \( M(2, 1, \frac{7}{2}) \) and radius \( r = \frac{9}{2} \): \[ \left(x - 2\right)^2 + \left(y - 1\right)^2 + \left(z - \frac{7}{2}\right)^2 = \left(\frac{9}{2}\right)^2. \] This simplifies to: \[ \left(x - 2\right)^2 + \left(y - 1\right)^2 + \left(z - \frac{7}{2}\right)^2 = \frac{81}{4}. \] Thus, the equation of the sphere is \[ \left(x - 2\right)^2 + \left(y - 1\right)^2 + \left(z - \frac{7}{2}\right)^2 = \frac{81}{4}. \]