3. Given four points $A, B, C, D$ on a sphere. $A B, A C, A D$ are mutually perpendicular, and $A B+A C+A D=$ 12, then the minimum value of the surface area of the sphere is
A. $40 \pi$
B. $45 \pi$
C. $48 \pi$
D. $52 \pi$
3. C Construct a rectangular parallelepiped with $A B, A C, A D$ as three edges starting from the same vertex, let $A B=a, A C=b, A D=c$, then the radius of the sphere is half the length of the diagonal of the rectangular parallelepiped, which is $\frac{1}{2} \sqrt{a^{2}+b^{2}+c^{2}}$.

So $S_{\text {sphere }}=4 \pi r^{2}=\pi\left(a^{2}+b^{2}+c^{2}\right) \geqslant \pi \cdot \frac{1}{3}(a+b+c)^{2}$ (Cauchy-Schwarz inequality)
$$
=48 \pi
$$

Equality holds if and only if $a=b=c=4$.