Let $\mathrm { A } ( x , y , z )$ be a point in $xy$-plane, which is equidistant from three points $( 0,3,2 ) , ( 2,0,3 )$ and $( 0,0,1 )$. Let $\mathrm { B } = ( 1,4 , - 1 )$ and $\mathrm { C } = ( 2,0 , - 2 )$. Then among the statements (S1) : $\triangle \mathrm { ABC }$ is an isosceles right angled triangle, and (S2) : the area of $\triangle \mathrm { ABC }$ is $\frac { 9 \sqrt { 2 } } { 2 }$,
(1) both are true
(2) only (S2) is true
(3) only (S1) is true
(4) both are false

Q79. Let the point, on the line passing through the points $P ( 1 , - 2,3 )$ and $Q ( 5 , - 4,7 )$, farther from the origin and at distance of 9 units from the point P , be $( \alpha , \beta , \gamma )$. Then $\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 }$ is equal to :
(1) 165
(2) 160
(3) 155
(4) 150

Suppose that a triangle ABC which is inscribed in a circle O of radius 2 satisfies
$$3\overrightarrow{\mathrm{OA}} + 4\overrightarrow{\mathrm{OB}} + 2\overrightarrow{\mathrm{OC}} = \vec{0}.$$
Let D denote the point of intersection of the straight line AO and the segment BC. We are to find the lengths of the segments AD and BD.
(1) When we set $\overrightarrow{\mathrm{OD}} = k\overrightarrow{\mathrm{OA}}$ where $k$ is a real number, we have
$$\overrightarrow{\mathrm{OD}} = -\frac{\mathbf{A}}{\mathbf{B}}k\overrightarrow{\mathrm{OB}} - \frac{\mathbf{C}}{\mathbf{D}}k\overrightarrow{\mathrm{OC}}.$$
As the three points B, C and D are located on a straight line, we obtain $k = \frac{\mathbf{EF}}{\mathbf{GH}}$. From this we derive that $\mathrm{OD} = \mathbf{H}$ and finally obtain
$$\mathrm{AD} = 1.$$
(2) From (1) we see that $\mathrm{BD} = \frac{\mathbf{J}}{\mathbf{K}}\mathrm{BC}$. So in order to find the length of the segment BD, we should find the length of the segment BC.
First we note that
$$\mathrm{BC}^2 = \mathbf{L} - \mathbf{M}\overrightarrow{\mathrm{OB}} \cdot \overrightarrow{\mathrm{OC}}$$
where $\overrightarrow{\mathrm{OB}} \cdot \overrightarrow{\mathrm{OC}}$ represents the inner product of $\overrightarrow{\mathrm{OB}}$ and $\overrightarrow{\mathrm{OC}}$. Since we know from (1) that $|4\overrightarrow{\mathrm{OB}} + 2\overrightarrow{\mathrm{OC}}|^2 = \mathbf{NO}$, we have
$$\overrightarrow{\mathrm{OB}} \cdot \overrightarrow{\mathrm{OC}} = \frac{\mathbf{PQR}}{\mathbf{S}}.$$
Hence we obtain $\mathrm{BC} = \frac{\square\sqrt{\mathbf{U}}}{\square\mathbf{V}}$ and finally from that
$$\mathrm{BD} = \frac{\sqrt{\mathrm{W}}}{\mathrm{X}}.$$

On the coordinate plane, take point $\mathrm { A } ( 2,0 )$, and on the circle centered at origin O with radius 2, take points $\mathrm { B } , \mathrm { C } , \mathrm { D } , \mathrm { E } , \mathrm { F }$ such that points $\mathrm { A } , \mathrm { B } , \mathrm { C } , \mathrm { D } , \mathrm { E } , \mathrm { F }$ are the vertices of a regular hexagon in order. Here, B is in the first quadrant.
(1) The coordinates of point B are (ア ア (ア $)$, and the coordinates of point D are $( -$ ウ, $0 )$.
(2) Let M be the midpoint of segment BD, and let N be the intersection of line AM and line CD. We want to find $\overrightarrow { \mathrm { ON } }$.
$\overrightarrow { \mathrm { ON } }$ can be expressed in two ways using real numbers $r, s$: $\overrightarrow { \mathrm { ON } } = \overrightarrow { \mathrm { OA } } + r \overrightarrow { \mathrm { AM } } , \overrightarrow { \mathrm { ON } } = \overright

In coordinate space, let $O$ be the origin, and let point $P$ be the intersection of three planes $x - 3y - 5z = 0$, $x - 3y + 2z = 0$, $x + y = t$, where $t > 0$. If $\overline{OP} = 10$, then $t =$ （9）（10）（11）. (Express as a simplified radical)

In $\triangle A B C$, $\overline { A B } = 6 , \overline { A C } = 5 , \overline { B C } = 4$. Let $D$ be the midpoint of $\overline { A B }$, and $P$ be the intersection of the angle bisector of $\angle A B C$ and $\overline { C D }$, as shown in the figure. Select the correct options.
(1) $\overline { C P } = \frac { 3 } { 7 } \overline { C D }$
(2) $\overrightarrow { A P } = \frac { 3 } { 7 } \overrightarrow { A B } + \frac { 2 } { 7 } \overrightarrow { A C }$
(3) $\cos \angle B A C = \frac { 3 } { 4 }$
(4) The area of $\triangle A C P$ is $\frac { 15 } { 14 } \sqrt { 7 }$
(5) (Dot product) $\overrightarrow { A P } \cdot \overrightarrow { A C } = \frac { 120 } { 7 }$