(Course 2) Q1 For $\mathbf { A } , \mathbf { B } , \mathbf { D } , \mathbf { E }$ and $\mathbf { G }$ in the following sentences, choose the correct answer from among choices (0) $\sim$ (9) below, and for the other $\square$, enter the correct number.

Given a sphere of radius 2 with the center at point O, we have a tetrahedron ABCD whose four vertices are on the sphere. Let $\mathrm { AB } = \mathrm { BC } = \mathrm { CA } = 2$ and side BD be a diameter of the sphere.

Set $\overrightarrow { \mathrm { OA } } = \vec { a } , \overrightarrow { \mathrm { OB } } = \vec { b }$ and $\overrightarrow { \mathrm { OC } } = \vec { c }$.\\
(1) Let M and N denote the midpoints of segments DA and BC, respectively. Then we have

$$\overrightarrow { \mathrm { DA } } = \mathbf { A } , \quad \overrightarrow { \mathrm { MN } } = \frac { \mathbf { B } } { \mathbf { C } } + \mathbf { D }$$

(2) When the midpoint of segment MN is denoted by P and the center of gravity of triangle BCD is denoted by G, we see that

$$\overrightarrow { \mathrm { OP } } = \frac { \mathbf { E } } { \mathbf { F } } , \quad \overrightarrow { \mathrm { OG } } = \frac { \mathbf { G } } { \mathbf { F } } , \quad | \overrightarrow { \mathrm { PG } } | = \frac { \sqrt { \mathbf { H } } } { \mathbf { I } }$$

Also, since $\overrightarrow { \mathrm { AG } } = \frac { \mathbf { K } } { \mathbf { L } } \overrightarrow { \mathrm { AP } }$, we see that the three points A, P and G are on a straight line.\\
(0) $\vec { a }$\\
(1) $\vec { b }$\\
(2) $\vec { c }$\\
(3) $\vec { a } - \vec { b }$\\
(4) $\vec { b } - \vec { c }$\\
(5) $\vec { c } - \vec { a }$\\
(6) $\vec { a } + \vec { b }$\\
(7) $\vec { b } + \vec { c }$\\
(8) $\vec { c } + \vec { a }$\\
(9) $\vec { a } + \vec { b } + \vec { c }$