Mix Examples-Co-ordinate Geometry of Three Dimensions Questions in English

(C) The vertices of the tetrahedron are $A(3,2,4)$,$B(x_1, y_1, 0)$,$C(x_2, y_2, 0)$,and $D(x_3, y_3, 0)$.
Triangle $BCD$ is formed by the intersection of lines $y=x$,$x+y=6$,and $y=1$.
Solving for the intersection points:
$1$) $y=x$ and $y=1 \Rightarrow x=1$. Vertex $B = (1,1,0)$.
$2$) $x+y=6$ and $y=1 \Rightarrow x=5$. Vertex $D = (5,1,0)$.
$3$) $y=x$ and $x+y=6 \Rightarrow 2x=6 \Rightarrow x=3, y=3$. Vertex $C = (3,3,0)$.
The centroid $G$ of a tetrahedron with vertices $(x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3), (x_4, y_4, z_4)$ is given by $\left(\frac{\sum x_i}{4}, \frac{\sum y_i}{4}, \frac{\sum z_i}{4}\right)$.
$G = \left(\frac{3+1+3+5}{4}, \frac{2+1+3+1}{4}, \frac{4+0+0+0}{4}\right) = \left(\frac{12}{4}, \frac{7}{4}, \frac{4}{4}\right) = \left(3, \frac{7}{4}, 1\right)$.

(C) Let $AD$ be the internal bisector of $\angle A$ meeting $BC$ at $D$.
According to the Angle Bisector Theorem,$D$ divides the side $BC$ in the ratio of the adjacent sides $AB:AC$.
First,calculate the lengths of sides $AB$ and $AC$:
$AB = \sqrt{(2-4)^2 + (3-7)^2 + (4-8)^2} = \sqrt{(-2)^2 + (-4)^2 + (-4)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$.
$AC = \sqrt{(2-4)^2 + (5-7)^2 + (7-8)^2} = \sqrt{(-2)^2 + (-2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
The ratio $AB:AC = 6:3 = 2:1$.
Using the section formula,the coordinates of $D$ are:
$D = \left( \frac{2(2) + 1(2)}{2+1}, \frac{2(5) + 1(3)}{2+1}, \frac{2(7) + 1(4)}{2+1} \right) = \left( \frac{4+2}{3}, \frac{10+3}{3}, \frac{14+4}{3} \right) = \left( 2, \frac{13}{3}, 6 \right)$.
Now,calculate the length of $AD$:
$AD = \sqrt{(2-4)^2 + (\frac{13}{3} - 7)^2 + (6-8)^2} = \sqrt{(-2)^2 + (-\frac{8}{3})^2 + (-2)^2} = \sqrt{4 + \frac{64}{9} + 4} = \sqrt{8 + \frac{64}{9}} = \sqrt{\frac{72+64}{9}} = \sqrt{\frac{136}{9}} = \frac{\sqrt{4 \times 34}}{3} = \frac{2\sqrt{34}}{3}$.

(B) Let the cube have side length $a$. Place the cube in a coordinate system such that its vertices are at $(0,0,0), (a,0,0), (0,a,0), (0,0,a), (a,a,0), (a,0,a), (0,a,a),$ and $(a,a,a)$.
Consider the diagonal of the cube starting from the origin $(0,0,0)$ to $(a,a,a)$. The vector representing this diagonal is $\vec{d_1} = a\hat{i} + a\hat{j} + a\hat{k}$.
Consider a face diagonal starting from the same origin $(0,0,0)$ to $(a,a,0)$. The vector representing this diagonal is $\vec{d_2} = a\hat{i} + a\hat{j} + 0\hat{k}$.
The angle $\theta$ between these two vectors is given by $\cos \theta = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}| |\vec{d_2}|}$.
Calculating the dot product: $\vec{d_1} \cdot \vec{d_2} = (a)(a) + (a)(a) + (a)(0) = 2a^2$.
Calculating the magnitudes: $|\vec{d_1}| = \sqrt{a^2 + a^2 + a^2} = a\sqrt{3}$ and $|\vec{d_2}| = \sqrt{a^2 + a^2 + 0} = a\sqrt{2}$.
Thus,$\cos \theta = \frac{2a^2}{(a\sqrt{3})(a\sqrt{2})} = \frac{2}{\sqrt{6}} = \frac{\sqrt{4}}{\sqrt{6}} = \sqrt{\frac{2}{3}}$.
Therefore,$\theta = \operatorname{Cos}^{-1}\left(\sqrt{\frac{2}{3}}\right)$.

(C) Let the vertices be $A(0,0,0)$,$B(3,4,0)$,and $C(0,12,5)$.
First,we calculate the lengths of the sides opposite to the vertices $A, B, C$ denoted by $a, b, c$ respectively.
$a = BC = \sqrt{(0-3)^2 + (12-4)^2 + (5-0)^2} = \sqrt{(-3)^2 + 8^2 + 5^2} = \sqrt{9 + 64 + 25} = \sqrt{98} = 7\sqrt{2}$.
$b = AC = \sqrt{(0-0)^2 + (12-0)^2 + (5-0)^2} = \sqrt{0^2 + 12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$.
$c = AB = \sqrt{(3-0)^2 + (4-0)^2 + (0-0)^2} = \sqrt{3^2 + 4^2 + 0^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
The $x$-coordinate of the incentre $I(x, y, z)$ is given by the formula $x = \frac{ax_A + bx_B + cx_C}{a+b+c}$.
Substituting the values: $x = \frac{(7\sqrt{2})(0) + (13)(3) + (5)(0)}{7\sqrt{2} + 13 + 5} = \frac{0 + 39 + 0}{18 + 7\sqrt{2}} = \frac{39}{18 + 7\sqrt{2}}$.

(A) First,calculate the side lengths of $\triangle ABC$:
$AB^2 = (2-0)^2 + (-1-1)^2 + (3-2)^2 = 4 + 4 + 1 = 9 \implies AB = 3$.
$BC^2 = (1-2)^2 + (-3+1)^2 + (1-3)^2 = 1 + 4 + 4 = 9 \implies BC = 3$.
$AC^2 = (1-0)^2 + (-3-1)^2 + (1-2)^2 = 1 + 16 + 1 = 18 \implies AC = 3\sqrt{2}$.
Since $AB^2 + BC^2 = 9 + 9 = 18 = AC^2$,the triangle is a right-angled triangle with the right angle at $B$.
In a right-angled triangle,the orthocentre $(H)$ is the vertex containing the right angle,so $H = B(2,-1,3)$.
The circumcentre $(O)$ is the midpoint of the hypotenuse $AC$.
$O = \left( \frac{0+1}{2}, \frac{1-3}{2}, \frac{2+1}{2} \right) = \left( \frac{1}{2}, -1, \frac{3}{2} \right)$.
The distance $OH$ is given by $\sqrt{(2 - 1/2)^2 + (-1 - (-1))^2 + (3 - 3/2)^2}$.
$OH = \sqrt{(3/2)^2 + 0^2 + (3/2)^2} = \sqrt{9/4 + 9/4} = \sqrt{18/4} = \frac{3\sqrt{2}}{2} = \frac{3}{\sqrt{2}}$.

(A) Let the point in the $xy$-plane be $P(x, y, 0)$.
Since $P$ is equidistant from $A(2, 0, 3)$,$B(0, 3, 2)$,and $C(0, 0, 1)$,we have $PA^2 = PB^2 = PC^2$.
$PA^2 = (x-2)^2 + (y-0)^2 + (0-3)^2 = x^2 - 4x + 4 + y^2 + 9 = x^2 + y^2 - 4x + 13$.
$PB^2 = (x-0)^2 + (y-3)^2 + (0-2)^2 = x^2 + y^2 - 6y + 9 + 4 = x^2 + y^2 - 6y + 13$.
$PC^2 = (x-0)^2 + (y-0)^2 + (0-1)^2 = x^2 + y^2 + 1$.
Equating $PA^2 = PC^2$: $x^2 + y^2 - 4x + 13 = x^2 + y^2 + 1 \implies -4x = -12 \implies x = 3$.
Equating $PB^2 = PC^2$: $x^2 + y^2 - 6y + 13 = x^2 + y^2 + 1 \implies -6y = -12 \implies y = 2$.
Thus,the coordinates of the point $P$ are $(3, 2, 0)$.

(B) Let the point be $P(x, y, z)$. The given points are $F_1(a, 0, 0)$ and $F_2(-a, 0, 0)$.
According to the problem,$PF_1 + PF_2 = 2k$.
Using the distance formula: $\sqrt{(x-a)^2 + y^2 + z^2} + \sqrt{(x+a)^2 + y^2 + z^2} = 2k$.
Let $d^2 = y^2 + z^2$. Then $\sqrt{(x-a)^2 + d^2} = 2k - \sqrt{(x+a)^2 + d^2}$.
Squaring both sides: $(x-a)^2 + d^2 = 4k^2 + (x+a)^2 + d^2 - 4k\sqrt{(x+a)^2 + d^2}$.
$x^2 - 2ax + a^2 = 4k^2 + x^2 + 2ax + a^2 - 4k\sqrt{(x+a)^2 + d^2}$.
$-4ax - 4k^2 = -4k\sqrt{(x+a)^2 + d^2}$.
$ax + k^2 = k\sqrt{(x+a)^2 + d^2}$.
Squaring again: $a^2x^2 + 2axk^2 + k^4 = k^2(x^2 + 2ax + a^2 + y^2 + z^2)$.
$a^2x^2 + 2axk^2 + k^4 = k^2x^2 + 2axk^2 + k^2a^2 + k^2(y^2 + z^2)$.
$x^2(a^2 - k^2) + k^2(y^2 + z^2) = k^2a^2 - k^4 = -k^2(k^2 - a^2)$.
Dividing by $-k^2(k^2 - a^2)$: $\frac{x^2(a^2 - k^2)}{-k^2(k^2 - a^2)} + \frac{k^2(y^2 + z^2)}{-k^2(k^2 - a^2)} = 1$.
$\frac{x^2}{k^2} + \frac{y^2 + z^2}{k^2 - a^2} = 1$.

(A) We know that in any triangle,the circumcentre $(S)$,centroid $(G)$,and orthocentre $(O)$ are collinear,and the centroid divides the line segment joining the circumcentre and the orthocentre in the ratio $1:2$.
Let the circumcentre be $S(x, y, z)$. Given $O(-3, 5, 2)$ and $G(3, 3, 4)$.
Using the section formula,the centroid $G$ is given by:
$G = \left(\frac{1 \cdot O + 2 \cdot S}{1+2}\right) = \left(\frac{-3+2x}{3}, \frac{5+2y}{3}, \frac{2+2z}{3}\right)$
Equating this to the given centroid $(3, 3, 4)$:
$\frac{-3+2x}{3} = 3 \Rightarrow -3+2x = 9 \Rightarrow 2x = 12 \Rightarrow x = 6$
$\frac{5+2y}{3} = 3 \Rightarrow 5+2y = 9 \Rightarrow 2y = 4 \Rightarrow y = 2$
$\frac{2+2z}{3} = 4 \Rightarrow 2+2z = 12 \Rightarrow 2z = 10 \Rightarrow z = 5$
Thus,the circumcentre is $(6, 2, 5)$.

(A) The centroid $G$ of a tetrahedron with vertices $A, B, C, D$ is given by $G = \left(\frac{x_1+x_2+x_3+x_4}{4}, \frac{y_1+y_2+y_3+y_4}{4}, \frac{z_1+z_2+z_3+z_4}{4}\right)$.
For the given points $A(3,2,-1), B(4,1,1), C(6,2,5), D(3,3,3)$,the centroid $G$ is:
$G = \left(\frac{3+4+6+3}{4}, \frac{2+1+2+3}{4}, \frac{-1+1+5+3}{4}\right) = \left(\frac{16}{4}, \frac{8}{4}, \frac{8}{4}\right) = (4, 2, 2)$.
The lines joining a vertex of a tetrahedron to the centroid of the opposite face are concurrent at the centroid of the tetrahedron.
Thus,the lines $AG_1, BG_2, CG_3$ and $DG_4$ are concurrent at the point $(4, 2, 2)$.

(D) Let the coordinates of the vertices of the tetrahedron be $A(a_1-d, a_1, a_1+d)$,$B(a_2-d, a_2, a_2+d)$,$C(a_3-d, a_3, a_3+d)$,and $D(a_4-d, a_4, a_4+d)$.
The centroid $G$ is given by the average of the coordinates of the vertices:
$G = \left(\frac{\sum a_i - 4d}{4}, \frac{\sum a_i}{4}, \frac{\sum a_i + 4d}{4}\right) = (2, 3, k)$.
Equating the coordinates,we get:
$1) \frac{\sum a_i - 4d}{4} = 2 \implies \sum a_i - 4d = 8$
$2) \frac{\sum a_i}{4} = 3 \implies \sum a_i = 12$
$3) \frac{\sum a_i + 4d}{4} = k \implies \sum a_i + 4d = 4k$
Substituting $\sum a_i = 12$ into the first equation: $12 - 4d = 8 \implies 4d = 4 \implies d = 1$.
Substituting $\sum a_i = 12$ and $d = 1$ into the third equation: $12 + 4(1) = 4k \implies 16 = 4k \implies k = 4$.
Thus,the centroid $G$ is $(2, 3, 4)$.
The distance of $G$ from the origin $(0, 0, 0)$ is $\sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$.

(D) The centroid of a tetrahedron with vertices $A, B, C, D$ is given by $G = \frac{A+B+C+D}{4}$.
Given $A(1,1,1), B(1,-4,3), C(2,-2,0), D(8,1,4)$.
The centroid of the tetrahedron $ABCD$ is $G = \left(\frac{1+1+2+8}{4}, \frac{1-4-2+1}{4}, \frac{1+3+0+4}{4}\right) = \left(\frac{12}{4}, \frac{-4}{4}, \frac{8}{4}\right) = (3,-1,2)$.
It is a known property that the centroid of the tetrahedron formed by the centroids of the faces of a tetrahedron is the same as the centroid of the original tetrahedron.
Therefore,the centroid of the tetrahedron with vertices $G_1, G_2, G_3, G_4$ is the same as the centroid of $ABCD$,which is $(3,-1,2)$.
Thus,option $D$ is correct.

(A) We know that the centroid of a triangle divides the line segment joining the orthocentre and the circumcentre in the ratio $2:1$.
Let the orthocentre be $O(5, 2, -6)$,the centroid be $G(9, 6, -4)$,and the circumcentre be $C(x, y, z)$.
Using the section formula,the centroid $G$ is given by:
$G = \left( \frac{2x + 1(5)}{2+1}, \frac{2y + 1(2)}{2+1}, \frac{2z + 1(-6)}{2+1} \right) = (9, 6, -4)$.
Equating the coordinates:
$9 = \frac{2x + 5}{3} \implies 27 = 2x + 5 \implies 2x = 22 \implies x = 11$.
$6 = \frac{2y + 2}{3} \implies 18 = 2y + 2 \implies 2y = 16 \implies y = 8$.
$-4 = \frac{2z - 6}{3} \implies -12 = 2z - 6 \implies 2z = -6 \implies z = -3$.
Therefore,the circumcentre is $(11, 8, -3)$.

(C) Let the points be $A(-4, 9, k)$,$B(-1, 6, k)$,and $C(0, 7, 10)$.
Calculate the squared distances between the points:
$AB^2 = (-1 - (-4))^2 + (6 - 9)^2 + (k - k)^2 = 3^2 + (-3)^2 + 0^2 = 9 + 9 = 18$.
$BC^2 = (0 - (-1))^2 + (7 - 6)^2 + (10 - k)^2 = 1^2 + 1^2 + (10 - k)^2 = 2 + (10 - k)^2$.
$AC^2 = (0 - (-4))^2 + (7 - 9)^2 + (10 - k)^2 = 4^2 + (-2)^2 + (10 - k)^2 = 16 + 4 + (10 - k)^2 = 20 + (10 - k)^2$.
For a right-angled isosceles triangle,two sides must be equal and the Pythagorean theorem must hold.
Case $1$: $AB = BC$.
$18 = 2 + (10 - k)^2 \implies (10 - k)^2 = 16 \implies 10 - k = \pm 4 \implies k = 6$ or $k = 14$.
If $k = 6$,$AB^2 = 18$,$BC^2 = 18$,$AC^2 = 20 + 16 = 36$. Since $18 + 18 = 36$,$AB^2 + BC^2 = AC^2$,so it is a right-angled isosceles triangle.
If $k = 14$,$AB^2 = 18$,$BC^2 = 18$,$AC^2 = 20 + 16 = 36$. Since $18 + 18 = 36$,it is a right-angled isosceles triangle.
Case $2$: $AB = AC$.
$18 = 20 + (10 - k)^2 \implies (10 - k)^2 = -2$,which is impossible.
Case $3$: $BC = AC$.
$2 + (10 - k)^2 = 20 + (10 - k)^2 \implies 2 = 20$,which is impossible.
Thus,there are $2$ possible values for $k$.

(D) Let the point be $P(x, y, z)$.
The distance of point $P(x, y, z)$ from the $XY$-plane is given by $|z|$.
The distance of point $P(x, y, z)$ from the $Z$-axis is given by $\sqrt{x^2 + y^2}$.
According to the problem,the distance from the $XY$-plane is twice the distance from the $Z$-axis:
$|z| = 2 \sqrt{x^2 + y^2}$.
Squaring both sides,we get:
$z^2 = 4(x^2 + y^2)$.
$z^2 = 4x^2 + 4y^2$.
Rearranging the terms,we get:
$4x^2 + 4y^2 - z^2 = 0$.
Thus,the correct option is $D$.

(B) The centroid $G$ of a tetrahedron with vertices $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,$C(x_3, y_3, z_3)$,and $D(x_4, y_4, z_4)$ is given by $\left(\frac{x_1+x_2+x_3+x_4}{4}, \frac{y_1+y_2+y_3+y_4}{4}, \frac{z_1+z_2+z_3+z_4}{4}\right)$.
For the given vertices $A(1, 2, 3)$,$B(-1, 4, 6)$,$C(0, -6, 4)$,and $D(1, 1, 1)$,the centroid $G$ is $\left(\frac{1-1+0+1}{4}, \frac{2+4-6+1}{4}, \frac{3+6+4+1}{4}\right) = \left(\frac{1}{4}, \frac{1}{4}, \frac{14}{4}\right)$.
The centroid $G_1$ of the face $BCD$ is given by $\left(\frac{x_2+x_3+x_4}{3}, \frac{y_2+y_3+y_4}{3}, \frac{z_2+z_3+z_4}{3}\right) = \left(\frac{-1+0+1}{3}, \frac{4-6+1}{3}, \frac{6+4+1}{3}\right) = \left(0, -\frac{1}{3}, \frac{11}{3}\right)$.
In a tetrahedron,the centroid $G$ divides the line segment joining a vertex to the centroid of the opposite face in the ratio $3:1$. Specifically,$G$ lies on the median $AG_1$ such that $AG : GG_1 = 3 : 1$.
This implies $AG = \frac{3}{4} AG_1$,or $\frac{AG_1}{AG} = \frac{4}{3}$.

(A) The centroid $G_1$ of triangle $ABC$ is given by $\left(\frac{1+2+0}{3}, \frac{2+0-2}{3}, \frac{0-1+3}{3}\right) = \left(1, 0, \frac{2}{3}\right)$.
The centroid $G_2$ of tetrahedron $ABCD$ is given by $\left(\frac{1+2+0-1}{4}, \frac{2+0-2+2}{4}, \frac{0-1+3-3}{4}\right) = \left(\frac{2}{4}, \frac{2}{4}, \frac{-1}{4}\right) = \left(\frac{1}{2}, \frac{1}{2}, -\frac{1}{4}\right)$.
Point $P$ divides $G_1G_2$ in the ratio $4:3$ internally. Using the section formula $\left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}, \frac{mz_2+nz_1}{m+n}\right)$ with $m=4, n=3$:
$x = \frac{4(1/2) + 3(1)}{4+3} = \frac{2+3}{7} = \frac{5}{7}$.
$y = \frac{4(1/2) + 3(0)}{4+3} = \frac{2+0}{7} = \frac{2}{7}$.
$z = \frac{4(-1/4) + 3(2/3)}{4+3} = \frac{-1+2}{7} = \frac{1}{7}$.
Thus,$P = \left(\frac{5}{7}, \frac{2}{7}, \frac{1}{7}\right)$.

(D) The centroid of a tetrahedron with vertices $(x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3)$ and $(x_4, y_4, z_4)$ is given by $\left(\frac{x_1+x_2+x_3+x_4}{4}, \frac{y_1+y_2+y_3+y_4}{4}, \frac{z_1+z_2+z_3+z_4}{4}\right)$.
Given vertices are $(a, 2, 1), (1, b, 4), (4, 0, c)$ and $(1, 1, 7)$.
So,the centroid is $\left(\frac{a+1+4+1}{4}, \frac{2+b+0+1}{4}, \frac{1+4+c+7}{4}\right) = \left(\frac{a+6}{4}, \frac{b+3}{4}, \frac{c+12}{4}\right)$.
Equating this to the given centroid $\left(\frac{9}{4}, \frac{5}{4}, \frac{15}{4}\right)$:
For $x$-coordinate: $\frac{a+6}{4} = \frac{9}{4} \Rightarrow a+6 = 9 \Rightarrow a = 3$.
For $y$-coordinate: $\frac{b+3}{4} = \frac{5}{4} \Rightarrow b+3 = 5 \Rightarrow b = 2$.
For $z$-coordinate: $\frac{c+12}{4} = \frac{15}{4} \Rightarrow c+12 = 15 \Rightarrow c = 3$.
Thus,$a=3, b=2, c=3$. Comparing these values,we get $a=c=b+1$ (since $3=3=2+1$).

(C) Point $E$ is the centroid of face $ABC$ (which is a triangle).
$\text{Centroid} = \left[\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right]$
$E = \left[\frac{2+(-3)+(-1)}{3}, \frac{3+3+4}{3}, \frac{-4+(-2)+2}{3}\right] = \left[\frac{-2}{3}, \frac{10}{3}, \frac{-4}{3}\right]$
Point $F$ is the centroid of face $ACD$.
$F = \left[\frac{2+(-1)+3}{3}, \frac{3+4+5}{3}, \frac{-4+2+1}{3}\right] = \left[\frac{4}{3}, \frac{12}{3}, \frac{-1}{3}\right] = \left[\frac{4}{3}, 4, \frac{-1}{3}\right]$
Point $G$ is the centroid of face $ABD$.
$G = \left[\frac{2+(-3)+3}{3}, \frac{3+3+5}{3}, \frac{-4+(-2)+1}{3}\right] = \left[\frac{2}{3}, \frac{11}{3}, \frac{-5}{3}\right]$
Now,$E, F, G$ form a triangle. The centroid of $\triangle EFG$ is given by the average of the coordinates of $E, F,$ and $G$.
$\text{Centroid of } \triangle EFG = \left[\frac{\frac{-2}{3}+\frac{4}{3}+\frac{2}{3}}{3}, \frac{\frac{10}{3}+4+\frac{11}{3}}{3}, \frac{\frac{-4}{3}+\left(\frac{-1}{3}\right)+\left(\frac{-5}{3}\right)}{3}\right]$
$= \left[\frac{\frac{4}{3}}{3}, \frac{\frac{10+12+11}{3}}{3}, \frac{\frac{-10}{3}}{3}\right] = \left[\frac{4}{9}, \frac{33}{9}, \frac{-10}{9}\right] = \left[\frac{4}{9}, \frac{11}{3}, \frac{-10}{9}\right]$

(C) In any triangle,the centroid divides the line segment joining the orthocentre and the circumcentre in the ratio $2: 1$.
Let the orthocentre be $O(-1, 3, 2)$ and the circumcentre be $C(5, 3, 2)$.
The centroid $G$ divides the line segment $OC$ in the ratio $2: 1$.
Using the section formula,the coordinates of $G$ are:
$G = \left( \frac{2(5) + 1(-1)}{2+1}, \frac{2(3) + 1(3)}{2+1}, \frac{2(2) + 1(2)}{2+1} \right)$
$G = \left( \frac{10-1}{3}, \frac{6+3}{3}, \frac{4+2}{3} \right)$
$G = \left( \frac{9}{3}, \frac{9}{3}, \frac{6}{3} \right) = (3, 3, 2)$.
Thus,Assertion $(A)$ is true.
Reason $(R)$ states that the ratio is $1: 2$,which is incorrect because the correct ratio is $2: 1$.
Therefore,$(A)$ is true and $(R)$ is false.

(C) Let the vertices of the cube be $(0,0,0), (a,0,0), (0,a,0), (0,0,a), (a,a,0), (a,0,a), (0,a,a), (a,a,a)$.
Consider two diagonals of the cube,for example,the vectors $\vec{d_1} = (a,a,a)$ and $\vec{d_2} = (-a,a,a)$.
The angle $\alpha$ between them is given by $\cos \alpha = \frac{\vec{d_1} \cdot \vec{d_2}}{|\vec{d_1}| |\vec{d_2}|} = \frac{-a^2+a^2+a^2}{3a^2} = \frac{1}{3}$.
Now,consider a diagonal of the cube $\vec{d_1} = (a,a,a)$ and a diagonal of a face that intersects it,for example,$\vec{f} = (a,a,0)$.
The angle $\beta$ between them is given by $\cos \beta = \frac{\vec{d_1} \cdot \vec{f}}{|\vec{d_1}| |\vec{f}|} = \frac{a^2+a^2+0}{\sqrt{3a^2} \sqrt{2a^2}} = \frac{2a^2}{a^2\sqrt{6}} = \sqrt{\frac{2}{3}}$.
Thus,$\cos^2 \beta = \frac{2}{3}$.
Finally,$\cos \alpha + \cos^2 \beta = \frac{1}{3} + \frac{2}{3} = 1$.

(A) The perpendicular distance of a point $P(x, y, z)$ from the $X$-axis is given by $d_1 = \sqrt{y^2 + z^2}$.
The perpendicular distance of the point $P(x, y, z)$ from the $YZ$-plane is given by $d_2 = |x|$.
According to the problem,the ratio of these distances is $d_1 : d_2 = 2 : 3$.
Therefore,$\frac{\sqrt{y^2 + z^2}}{|x|} = \frac{2}{3}$.
Squaring both sides,we get $\frac{y^2 + z^2}{x^2} = \frac{4}{9}$.
Cross-multiplying,we obtain $9(y^2 + z^2) = 4x^2$.
Rearranging the terms,we get $4x^2 - 9y^2 - 9z^2 = 0$.

(D) The coordinates of point $A$ are $(x, y, z) = (27, -243, 81)$.
The image of a point $(x, y, z)$ with respect to the $XY$-plane is $(x, y, -z)$. Thus,$B = (27, -243, -81)$.
The image of a point $(x, y, z)$ with respect to the $YZ$-plane is $(-x, y, z)$. Thus,$C = (-27, -243, 81)$.
The image of a point $(x, y, z)$ with respect to the $ZX$-plane is $(x, -y, z)$. Thus,$D = (27, 243, 81)$.
The centroid $(\alpha, \beta, \gamma)$ of triangle $BCD$ is given by the average of the coordinates of its vertices:
$\alpha = \frac{27 - 27 + 27}{3} = \frac{27}{3} = 9$
$\beta = \frac{-243 - 243 + 243}{3} = \frac{-243}{3} = -81$
$\gamma = \frac{-81 + 81 + 81}{3} = \frac{81}{3} = 27$
Therefore,$\alpha + \beta + \gamma = 9 - 81 + 27 = -45$.

(A) Let the origin be $O(0, 0, 0)$. The vertices of the rectangular parallelopiped are $O(0, 0, 0)$,$A(a, 0, 0)$,$E(0, b, 0)$,$D(0, 0, c)$,and others.
$d_1$ is a diagonal of the face in the $XY$-plane not passing through the origin. The vertices of this face are $(0, 0, 0), (a, 0, 0), (0, b, 0), (a, b, 0)$. The diagonal not passing through the origin is the line segment connecting $(a, 0, 0)$ and $(0, b, 0)$,i.e.,$AE$.
The vector along $d_1$ is $\vec{v_1} = (0-a)\hat{i} + (b-0)\hat{j} + (0-0)\hat{k} = -a\hat{i} + b\hat{j}$.
$d_2$ is a diagonal of the plane $\Pi_2$ (which is parallel to $ZX$ plane at distance $b$) coterminous with $d_1$. The plane $\Pi_2$ contains points $(0, b, 0), (a, b, 0), (0, b, c), (a, b, c)$. The diagonal coterminous with $d_1$ (which starts at $A(a, 0, 0)$) is $AD$,where $D$ is $(0, 0, c)$.
Wait,looking at the provided image,$d_1$ is $AE$ and $d_2$ is $AD$.
The vector $\vec{AE} = (0-a)\hat{i} + (b-0)\hat{j} + (0-0)\hat{k} = -a\hat{i} + b\hat{j}$.
The vector $\vec{AD} = (0-a)\hat{i} + (0-0)\hat{j} + (c-0)\hat{k} = -a\hat{i} + c\hat{k}$.
The angle $\theta$ between them is given by $\cos \theta = \frac{\vec{AE} \cdot \vec{AD}}{|\vec{AE}| |\vec{AD}|}$.
$\vec{AE} \cdot \vec{AD} = (-a)(-a) + (b)(0) + (0)(c) = a^2$.
$|\vec{AE}| = \sqrt{(-a)^2 + b^2} = \sqrt{a^2+b^2}$.
$|\vec{AD}| = \sqrt{(-a)^2 + c^2} = \sqrt{a^2+c^2}$.
Thus,$\cos \theta = \frac{a^2}{\sqrt{a^2+b^2} \sqrt{a^2+c^2}}$.
Therefore,$\theta = \cos ^{-1}\left(\frac{a^2}{\sqrt{a^2+b^2} \sqrt{a^2+c^2}}\right)$.