Let C be the circle formed by the intersection of the sphere $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 4$ and the plane $z = - 1$. When a plane $\alpha$ containing the $x$-axis intersects the sphere $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 4$ to form a circle that meets C at exactly one point, and a normal vector to plane $\alpha$ is $\vec { n } = ( a , 3 , b )$, find the value of $a ^ { 2 } + b ^ { 2 }$. [4 points]

In coordinate space, there is a circle $C$ formed by the intersection of the sphere $x ^ { 2 } + y ^ { 2 } + z ^ { 2 } = 6$ and the plane $x + 2 z - 5 = 0$. Let P be the point on circle $C$ with the minimum $y$-coordinate, and let Q be the foot of the perpendicular from point P to the $xy$-plane. For a point X moving on circle $C$, the maximum value of $| \overrightarrow { \mathrm { PX } } + \overrightarrow { \mathrm { QX } } | ^ { 2 }$ is $a + b \sqrt { 30 }$.
Find the value of $10 ( a + b )$. (Here, $a$ and $b$ are rational numbers.) [4 points]

As shown in the figure, for a tetrahedron ABCD with $\overline{\mathrm{AB}} = 6$, $\overline{\mathrm{BC}} = 4\sqrt{5}$, let M be the midpoint of segment BC. Triangle AMD is equilateral and line BC is perpendicular to plane AMD. Find the area of the orthogonal projection of the circle inscribed in triangle ACD onto plane BCD. [3 points]
(1) $\frac{\sqrt{10}}{4}\pi$
(2) $\frac{\sqrt{10}}{6}\pi$
(3) $\frac{\sqrt{10}}{8}\pi$
(4) $\frac{\sqrt{10}}{10}\pi$
(5) $\frac{\sqrt{10}}{12}\pi$