Triangle and Parallelogram Questions in English

(D) Let the coordinates of vertex $B$ be $(x_1, y_1, z_1)$.
The centroid $G$ of a triangle with vertices $A(x_A, y_A, z_A)$,$B(x_B, y_B, z_B)$,and $C(x_C, y_C, z_C)$ is given by $\left(\frac{x_A+x_B+x_C}{3}, \frac{y_A+y_B+y_C}{3}, \frac{z_A+z_B+z_C}{3}\right)$.
Given $A(1,4,2)$,$C(5,-7,1)$,and $G\left(\frac{4}{3}, 0, \frac{-2}{3}\right)$,we have:
$\frac{1+x_1+5}{3} = \frac{4}{3} \Rightarrow 6+x_1 = 4 \Rightarrow x_1 = -2$
$\frac{4+y_1-7}{3} = 0 \Rightarrow y_1-3 = 0 \Rightarrow y_1 = 3$
$\frac{2+z_1+1}{3} = \frac{-2}{3} \Rightarrow 3+z_1 = -2 \Rightarrow z_1 = -5$
So,$B = (-2, 3, -5)$.
The midpoint of side $BC$ is $\left(\frac{x_B+x_C}{2}, \frac{y_B+y_C}{2}, \frac{z_B+z_C}{2}\right)$.
Midpoint $= \left(\frac{-2+5}{2}, \frac{3-7}{2}, \frac{-5+1}{2}\right) = \left(\frac{3}{2}, -2, -2\right)$.

(D) The centroid $G$ of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3})$.
Given $A(7, -8, 1)$,$B(p, q, 5)$,$C(q+1, 5p, 0)$,and $G(3, -5, r)$.
Equating the coordinates:
$x$-coordinate: $\frac{7+p+q+1}{3} = 3 \Rightarrow p+q+8 = 9 \Rightarrow p+q = 1$ (Equation $1$)
$y$-coordinate: $\frac{-8+q+5p}{3} = -5 \Rightarrow 5p+q-8 = -15 \Rightarrow 5p+q = -7$ (Equation $2$)
$z$-coordinate: $\frac{1+5+0}{3} = r \Rightarrow r = \frac{6}{3} = 2$
Subtracting Equation $1$ from Equation $2$: $(5p+q) - (p+q) = -7 - 1 \Rightarrow 4p = -8 \Rightarrow p = -2$.
Substituting $p = -2$ into Equation $1$: $-2 + q = 1 \Rightarrow q = 3$.
Thus,the values are $p = -2, q = 3, r = 2$.

(D) Let $D$ be the midpoint of $BC$. The coordinates of $D$ are given by:
$D = \left( \frac{\lambda - 1}{2}, \frac{5 + 3}{2}, \frac{\mu + 2}{2} \right) = \left( \frac{\lambda - 1}{2}, 4, \frac{\mu + 2}{2} \right)$.
The direction ratios of the median $AD$ are given by the difference of the coordinates of $D$ and $A$:
$AD = \left( \frac{\lambda - 1}{2} - 2, 4 - 3, \frac{\mu + 2}{2} - 5 \right) = \left( \frac{\lambda - 5}{2}, 1, \frac{\mu - 8}{2} \right)$.
Since the median $AD$ is equally inclined to the coordinate axes,its direction ratios must be proportional to $(1, 1, 1)$ or $(\pm 1, \pm 1, \pm 1)$.
Thus,we have:
$\frac{\lambda - 5}{2} = 1 \implies \lambda - 5 = 2 \implies \lambda = 7$.
$\frac{\mu - 8}{2} = 1 \implies \mu - 8 = 2 \implies \mu = 10$.
Therefore,the values are $\lambda = 7$ and $\mu = 10$.

(D) Given vertices are $A \equiv(0,3,0), B \equiv(0,0,4)$,and $C \equiv(0,3,4)$.
First,calculate the lengths of the sides:
$a = BC = \sqrt{(0-0)^2 + (3-0)^2 + (4-4)^2} = \sqrt{0+9+0} = 3$
$b = CA = \sqrt{(0-0)^2 + (3-3)^2 + (4-0)^2} = \sqrt{0+0+16} = 4$
$c = AB = \sqrt{(0-0)^2 + (0-3)^2 + (4-0)^2} = \sqrt{0+9+16} = \sqrt{25} = 5$
Now,the Incenter $I$ is given by $\left(\frac{ax_1+bx_2+cx_3}{a+b+c}, \frac{ay_1+by_2+cy_3}{a+b+c}, \frac{az_1+bz_2+cz_3}{a+b+c}\right)$:
$I = \left(\frac{3(0)+4(0)+5(0)}{3+4+5}, \frac{3(3)+4(0)+5(3)}{3+4+5}, \frac{3(0)+4(4)+5(4)}{3+4+5}\right)$
$I = \left(\frac{0}{12}, \frac{9+0+15}{12}, \frac{0+16+20}{12}\right) = \left(0, \frac{24}{12}, \frac{36}{12}\right) = (0, 2, 3)$
The Centroid $G$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$:
$G = \left(\frac{0+0+0}{3}, \frac{3+0+3}{3}, \frac{0+4+4}{3}\right) = \left(0, \frac{6}{3}, \frac{8}{3}\right) = \left(0, 2, \frac{8}{3}\right)$
Thus,the incenter and centroid are $(0, 2, 3)$ and $\left(0, 2, \frac{8}{3}\right)$ respectively.

(C) In $\triangle ABP$ and $\triangle CDP$,$AB \parallel DC$ and $AB = DC$. Since $P$ is the mid-point of $AB$,$AP = PB = \frac{1}{2} AB = \frac{1}{2} DC$.
Consider $\triangle APR$ and $\triangle CPD$:
$\angle PAR = \angle PCD$ (alternate interior angles as $AB \parallel DC$)
$\angle APR = \angle CPD$ (vertically opposite angles)
Thus,$\triangle APR \sim \triangle CPD$ by $AA$ similarity.
Therefore,the ratio of corresponding sides is equal:
$\frac{AR}{CR} = \frac{AP}{CD} = \frac{\frac{1}{2} AB}{AB} = \frac{1}{2}$.
Hence,$R$ divides $AC$ in the ratio $1: 2$.

(A) Given the vertices of the triangle are $P(3,2,6), Q(1,4,5)$,and $R(3,5,3)$.
First,we find the direction ratios of the vectors $\vec{QP}$ and $\vec{QR}$.
The vector $\vec{QP} = (3-1, 2-4, 6-5) = (2, -2, 1)$.
The vector $\vec{QR} = (3-1, 5-4, 3-5) = (2, 1, -2)$.
Now,we calculate the dot product of $\vec{QP}$ and $\vec{QR}$:
$\vec{QP} \cdot \vec{QR} = (2)(2) + (-2)(1) + (1)(-2) = 4 - 2 - 2 = 0$.
Since the dot product is $0$,the vectors $\vec{QP}$ and $\vec{QR}$ are perpendicular to each other.
Therefore,$m \angle PQR = 90^{\circ}$.

(A) The centroid $(G)$ of a tetrahedron with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,$(x_3, y_3, z_3)$,and $(x_4, y_4, z_4)$ is given by the formula:
$G = \left(\frac{x_1+x_2+x_3+x_4}{4}, \frac{y_1+y_2+y_3+y_4}{4}, \frac{z_1+z_2+z_3+z_4}{4}\right)$
Substituting the given coordinates $A(-1, 2, 3)$,$B(3, -2, 1)$,$C(2, 1, 3)$,and $D(-1, -2, 4)$:
$G = \left(\frac{-1+3+2-1}{4}, \frac{2-2+1-2}{4}, \frac{3+1+3+4}{4}\right)$
$G = \left(\frac{3}{4}, \frac{-1}{4}, \frac{11}{4}\right)$
Thus,the correct option is $A$.

(B) The centroid of a triangle formed by joining the midpoints of the sides of another triangle is the same as the centroid of the original triangle.
Let the midpoints be $M_1 = (1, 5, -1)$,$M_2 = (0, 4, -2)$,and $M_3 = (2, 3, 4)$.
The centroid $G(x, y, z)$ of the triangle formed by these midpoints is given by the average of their coordinates:
$x = \frac{1 + 0 + 2}{3} = \frac{3}{3} = 1$
$y = \frac{5 + 4 + 3}{3} = \frac{12}{3} = 4$
$z = \frac{-1 - 2 + 4}{3} = \frac{1}{3}$
Therefore,the centroid is $(1, 4, 1/3)$.

(D) The vertices of the triangle are $O(0,0,0)$,$A(3,0,0)$,and $B(0,4,0)$.
Let the side lengths be $a, b, c$ opposite to vertices $A, B, O$ respectively.
$a = |\overrightarrow{AB}| = \sqrt{(0-3)^2 + (4-0)^2 + (0-0)^2} = \sqrt{9+16} = 5$.
$b = |\overrightarrow{OB}| = \sqrt{(0-0)^2 + (4-0)^2 + (0-0)^2} = 4$.
$c = |\overrightarrow{OA}| = \sqrt{(3-0)^2 + (0-0)^2 + (0-0)^2} = 3$.
The coordinates of the incentre $I$ are given by $\left( \frac{ax_1 + bx_2 + cx_3}{a+b+c}, \frac{ay_1 + by_2 + cy_3}{a+b+c}, \frac{az_1 + bz_2 + cz_3}{a+b+c} \right)$.
Substituting the values: $I = \left( \frac{5(0) + 4(3) + 3(0)}{5+4+3}, \frac{5(0) + 4(0) + 3(4)}{5+4+3}, \frac{5(0) + 4(0) + 3(0)}{5+4+3} \right)$.
$I = \left( \frac{12}{12}, \frac{12}{12}, \frac{0}{12} \right) = (1, 1, 0)$.

(C) The centroid of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
Given the vertices $(a, 1, 3)$,$(-2, b, -5)$,and $(4, 7, c)$ and the centroid as $(0, 0, 0)$:
$\frac{a-2+4}{3} = 0$ $\Rightarrow a+2 = 0$ $\Rightarrow a = -2$
$\frac{1+b+7}{3} = 0$ $\Rightarrow b+8 = 0$ $\Rightarrow b = -8$
$\frac{3-5+c}{3} = 0$ $\Rightarrow c-2 = 0$ $\Rightarrow c = 2$
Therefore,$a^2 + b^2 + c^2 = (-2)^2 + (-8)^2 + (2)^2 = 4 + 64 + 4 = 72$.

(C) Let the coordinates of the vertices be $O(0, 0, 0)$,$A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$.
The centroid $G$ of the tetrahedron $OABC$ is given by $G = \left(\frac{0+x_1+x_2+x_3}{4}, \frac{0+y_1+y_2+y_3}{4}, \frac{0+z_1+z_2+z_3}{4}\right) = (2, -1, 2)$.
This implies $\frac{x_1+x_2+x_3}{4} = 2 \Rightarrow x_1+x_2+x_3 = 8$,$\frac{y_1+y_2+y_3}{4} = -1 \Rightarrow y_1+y_2+y_3 = -4$,and $\frac{z_1+z_2+z_3}{4} = 2 \Rightarrow z_1+z_2+z_3 = 8$.
The centroid $G_1$ of $\triangle ABC$ is given by $G_1 = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
Substituting the values,we get $G_1 = \left(\frac{8}{3}, \frac{-4}{3}, \frac{8}{3}\right)$.
The distance $\left|\overline{O G_1}\right| = \sqrt{\left(\frac{8}{3}\right)^2 + \left(-\frac{4}{3}\right)^2 + \left(\frac{8}{3}\right)^2} = \sqrt{\frac{64}{9} + \frac{16}{9} + \frac{64}{9}} = \sqrt{\frac{144}{9}} = \sqrt{16} = 4$.

(A) Let the vertices of the triangle be $A(0,0,0), B(3,0,0)$,and $C(0,4,0)$.
First,we calculate the lengths of the sides opposite to the vertices $A, B$,and $C$:
$a = BC = \sqrt{(0-3)^2 + (4-0)^2 + (0-0)^2} = \sqrt{(-3)^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5$
$b = AC = \sqrt{(0-0)^2 + (4-0)^2 + (0-0)^2} = \sqrt{4^2} = 4$
$c = AB = \sqrt{(3-0)^2 + (0-0)^2 + (0-0)^2} = \sqrt{3^2} = 3$
The coordinates of the incenter $(x, y, z)$ are given by the formula:
$x = \frac{ax_1 + bx_2 + cx_3}{a+b+c}, y = \frac{ay_1 + by_2 + cy_3}{a+b+c}, z = \frac{az_1 + bz_2 + cz_3}{a+b+c}$
Substituting the values $A(x_1, y_1, z_1) = (0,0,0), B(x_2, y_2, z_2) = (3,0,0), C(x_3, y_3, z_3) = (0,4,0)$ and side lengths $a=5, b=4, c=3$:
$x = \frac{5(0) + 4(3) + 3(0)}{5+4+3} = \frac{12}{12} = 1$
$y = \frac{5(0) + 4(0) + 3(4)}{5+4+3} = \frac{12}{12} = 1$
$z = \frac{5(0) + 4(0) + 3(0)}{5+4+3} = \frac{0}{12} = 0$
Thus,the incenter is $(1, 1, 0)$.

(A) Given points are $P(0, 7, 10)$,$Q(-1, 6, 6)$,and $R(-4, 9, 6)$.
Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$:
$PQ = \sqrt{(-1-0)^2 + (6-7)^2 + (6-10)^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}$.
$QR = \sqrt{(-4 - (-1))^2 + (9-6)^2 + (6-6)^2} = \sqrt{(-3)^2 + 3^2 + 0^2} = \sqrt{9 + 9 + 0} = \sqrt{18} = 3\sqrt{2}$.
$PR = \sqrt{(-4-0)^2 + (9-7)^2 + (6-10)^2} = \sqrt{(-4)^2 + 2^2 + (-4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$.
Since $PQ = QR = 3\sqrt{2}$,the triangle is isosceles.
Also,$PQ^2 + QR^2 = 18 + 18 = 36 = PR^2$. By the converse of the Pythagorean theorem,$\triangle PQR$ is a right-angled triangle.
Therefore,$\triangle PQR$ is a right-angled isosceles triangle.

(A) The centroid of $\triangle ABC$ is $G = \left(\frac{3+4+6}{3}, \frac{2+1+2}{3}, \frac{-1+1+5}{3}\right) = \left(\frac{13}{3}, \frac{5}{3}, \frac{5}{3}\right)$.
Since $D, E, F$ divide the sides $BC, CA, AB$ in the same ratio $k:1$ (where $k=2$),the centroid of $\triangle DEF$ is the same as the centroid of $\triangle ABC$.
Therefore,the centroid of $\triangle DEF$ is $\left(\frac{13}{3}, \frac{5}{3}, \frac{5}{3}\right)$.

(D) Let $A(1, 2, 3), B(3, -1, 5), C(4, 0, -3)$ be the vertices of the triangle. Let $O(x, y, z)$ be the circumcentre. Then $OA = OB = OC$,which implies $OA^2 = OB^2 = OC^2$.
$OA^2 = OB^2 \Rightarrow (x-1)^2 + (y-2)^2 + (z-3)^2 = (x-3)^2 + (y+1)^2 + (z-5)^2$
$x^2 - 2x + 1 + y^2 - 4y + 4 + z^2 - 6z + 9 = x^2 - 6x + 9 + y^2 + 2y + 1 + z^2 - 10z + 25$
$4x - 6y + 4z = 20 \Rightarrow 2x - 3y + 2z = 10 \quad \dots (i)$
$OB^2 = OC^2 \Rightarrow (x-3)^2 + (y+1)^2 + (z-5)^2 = (x-4)^2 + y^2 + (z+3)^2$
$x^2 - 6x + 9 + y^2 + 2y + 1 + z^2 - 10z + 25 = x^2 - 8x + 16 + y^2 + z^2 + 6z + 9$
$2x + 2y - 16z = -10 \Rightarrow x + y - 8z = -5 \quad \dots (ii)$
$OA^2 = OC^2 \Rightarrow (x-1)^2 + (y-2)^2 + (z-3)^2 = (x-4)^2 + y^2 + (z+3)^2$
$x^2 - 2x + 1 + y^2 - 4y + 4 + z^2 - 6z + 9 = x^2 - 8x + 16 + y^2 + z^2 + 6z + 9$
$6x - 4y - 12z = 11 \quad \dots (iii)$
Solving equations $(i), (ii),$ and $(iii)$,we get $x = \frac{7}{2}, y = -\frac{1}{2}, z = 1$.
Thus,the circumcentre is $\left(\frac{7}{2}, -\frac{1}{2}, 1\right)$.

(A) The centroid $G$ of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3})$.
For vertices $(2, 3, -1)$,$(5, 6, 3)$,$(2, -3, 1)$:
$G = (\frac{2+5+2}{3}, \frac{3+6-3}{3}, \frac{-1+3+1}{3}) = (\frac{9}{3}, \frac{6}{3}, \frac{3}{3}) = (3, 2, 1)$. Thus,$A-s$.
$(B)$ For an equilateral triangle,the circumcentre is the same as the centroid.
For vertices $(1, 2, 3)$,$(2, 3, 1)$,$(3, 1, 2)$:
$C = (\frac{1+2+3}{3}, \frac{2+3+1}{3}, \frac{3+1+2}{3}) = (\frac{6}{3}, \frac{6}{3}, \frac{6}{3}) = (2, 2, 2)$. Thus,$B-p$.
$(C)$ For an equilateral triangle,the orthocentre is the same as the centroid.
For vertices $(2, 1, 5)$,$(3, 2, 3)$,$(4, 0, 4)$:
$O = (\frac{2+3+4}{3}, \frac{1+2+0}{3}, \frac{5+3+4}{3}) = (\frac{9}{3}, \frac{3}{3}, \frac{12}{3}) = (3, 1, 4)$. Thus,$C-q$.
$(D)$ For a right-angled triangle with vertices at $(0, 0, 0)$,$(a, 0, 0)$,and $(0, b, 0)$,the incentre is $(\frac{ab}{a+b+\sqrt{a^2+b^2}}, \frac{ab}{a+b+\sqrt{a^2+b^2}}, 0)$.
Here,vertices are $(0, 0, 0)$,$(3, 0, 0)$,$(0, 4, 0)$. So,$a=3, b=4$. The hypotenuse $c = \sqrt{3^2+4^2} = 5$.
$I = (\frac{3 \times 4}{3+4+5}, \frac{3 \times 4}{3+4+5}, 0) = (\frac{12}{12}, \frac{12}{12}, 0) = (1, 1, 0)$. Thus,$D-r$.

(A) The centroid $G$ of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
Given $G = \left(5, -1, \frac{2}{3}\right)$,we have:
$\frac{1+h-4}{3} = 5$ $\Rightarrow h-3 = 15$ $\Rightarrow h = 18$.
$\frac{2-3+k}{3} = -1$ $\Rightarrow k-1 = -3$ $\Rightarrow k = -2$.
Thus,the vertices are $A(1, 2, 3)$,$B(18, -3, 0)$,and $C(-4, -2, -1)$.
Calculate the squares of the side lengths:
$AB^2 = (18-1)^2 + (-3-2)^2 + (0-3)^2 = 17^2 + (-5)^2 + (-3)^2 = 289 + 25 + 9 = 323$.
$BC^2 = (-4-18)^2 + (-2-(-3))^2 + (-1-0)^2 = (-22)^2 + 1^2 + (-1)^2 = 484 + 1 + 1 = 486$.
$CA^2 = (-4-1)^2 + (-2-2)^2 + (-1-3)^2 = (-5)^2 + (-4)^2 + (-4)^2 = 25 + 16 + 16 = 57$.
Since $BC^2 = 486$ and $AB^2 + CA^2 = 323 + 57 = 380$,we observe that $BC^2 > AB^2 + CA^2$.
Therefore,the triangle is an obtuse angled triangle.

(A) Let the points be $A(2, 3, 4)$,$B(-1, -2, 1)$,and $C(5, 8, 7)$.
We check if the points are collinear by calculating the distance between them.
$AB = \sqrt{(-1-2)^2 + (-2-3)^2 + (1-4)^2} = \sqrt{(-3)^2 + (-5)^2 + (-3)^2} = \sqrt{9 + 25 + 9} = \sqrt{43}$.
$BC = \sqrt{(5-(-1))^2 + (8-(-2))^2 + (7-1)^2} = \sqrt{6^2 + 10^2 + 6^2} = \sqrt{36 + 100 + 36} = \sqrt{172} = 2\sqrt{43}$.
$AC = \sqrt{(5-2)^2 + (8-3)^2 + (7-4)^2} = \sqrt{3^2 + 5^2 + 3^2} = \sqrt{9 + 25 + 9} = \sqrt{43}$.
Since $AB + AC = \sqrt{43} + \sqrt{43} = 2\sqrt{43} = BC$,the points $A, B, C$ are collinear.

(C) Given vertices of the parallelogram $PQRS$ are $P(1, 2)$,$Q(4, 6)$,$R(5, 7)$,and $S(a, b)$.
In a parallelogram,the diagonals bisect each other,meaning they share the same midpoint.
Let $A$ be the midpoint of the diagonals $PR$ and $QS$.
The midpoint $A$ of diagonal $PR$ is given by:
$A = \left( \frac{1 + 5}{2}, \frac{2 + 7}{2} \right) = \left( \frac{6}{2}, \frac{9}{2} \right) = \left( 3, 4.5 \right)$.
The midpoint $A$ of diagonal $QS$ is given by:
$A = \left( \frac{4 + a}{2}, \frac{6 + b}{2} \right)$.
Equating the coordinates of the midpoints:
$\frac{4 + a}{2} = 3 \implies 4 + a = 6 \implies a = 2$.
$\frac{6 + b}{2} = 4.5 \implies 6 + b = 9 \implies b = 3$.
Therefore,the values are $a = 2$ and $b = 3$.

(A) Let the vertices of the parallelogram be $A(2,4,-1)$,$B(3,6,-1)$,and $C(4,5,1)$.
Let the fourth vertex be $D(x, y, z)$.
In a parallelogram,the diagonals bisect each other,which means the midpoint of diagonal $AC$ is the same as the midpoint of diagonal $BD$.
Midpoint of $AC = \left( \frac{2+4}{2}, \frac{4+5}{2}, \frac{-1+1}{2} \right) = \left( 3, \frac{9}{2}, 0 \right)$.
Midpoint of $BD = \left( \frac{3+x}{2}, \frac{6+y}{2}, \frac{-1+z}{2} \right)$.
Equating the midpoints:
$\frac{3+x}{2} = 3 \Rightarrow 3+x = 6 \Rightarrow x = 3$.
$\frac{6+y}{2} = \frac{9}{2} \Rightarrow 6+y = 9 \Rightarrow y = 3$.
$\frac{-1+z}{2} = 0 \Rightarrow -1+z = 0 \Rightarrow z = 1$.
Thus,the fourth vertex $D$ is $(3,3,1)$.

(B) The vertices of $\triangle ABC$ are $A(2, -1, 1)$,$B(1, -3, -5)$,and $C(3, -4, -4)$.
First,we calculate the lengths of the sides using the distance formula:
$AB = \sqrt{(1-2)^2 + (-3 - (-1))^2 + (-5-1)^2} = \sqrt{(-1)^2 + (-2)^2 + (-6)^2} = \sqrt{1 + 4 + 36} = \sqrt{41}$.
$BC = \sqrt{(3-1)^2 + (-4 - (-3))^2 + (-4 - (-5))^2} = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.
$CA = \sqrt{(2-3)^2 + (-1 - (-4))^2 + (1 - (-4))^2} = \sqrt{(-1)^2 + 3^2 + 5^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$.
Now,check for the Pythagorean theorem:
$BC^2 + CA^2 = 6 + 35 = 41 = AB^2$.
Since $AB^2 = BC^2 + CA^2$,the triangle satisfies the condition for a right-angled triangle.
Thus,$\triangle ABC$ is a right-angled triangle.

(D) Let the coordinates of $C$ be $(x, y, z)$.
Since $G(1,0,1)$ is the centroid of $\triangle ABC$,we have:
$\frac{1+3+x}{3} = 1 \implies 4+x = 3 \implies x = -1$
$\frac{-4+1+y}{3} = 0 \implies -3+y = 0 \implies y = 3$
$\frac{2+0+z}{3} = 1 \implies 2+z = 3 \implies z = 1$
So,$C = (-1, 3, 1)$.
Now,calculate $AG^2$:
$AG^2 = (1-1)^2 + (0-(-4))^2 + (1-2)^2 = 0^2 + 4^2 + (-1)^2 = 16 + 1 = 17$.
Calculate $CG^2$:
$CG^2 = (-1-1)^2 + (3-0)^2 + (1-1)^2 = (-2)^2 + 3^2 + 0^2 = 4 + 9 = 13$.
Thus,$AG^2 + CG^2 = 17 + 13 = 30$.
Calculate $BG^2$:
$BG^2 = (3-1)^2 + (1-0)^2 + (0-1)^2 = 2^2 + 1^2 + (-1)^2 = 4 + 1 + 1 = 6$.
Comparing the values,$AG^2 + CG^2 = 30 = 5 \times 6 = 5 BG^2$.

(D) To determine the nature of the triangle formed by points $A(-1, 2, 3)$,$B(2, -3, 1)$,and $C(3, 1, -2)$,we calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.
$1$. Length of $AB$:
$AB = \sqrt{(2 - (-1))^2 + (-3 - 2)^2 + (1 - 3)^2} = \sqrt{3^2 + (-5)^2 + (-2)^2} = \sqrt{9 + 25 + 4} = \sqrt{38}$.
$2$. Length of $BC$:
$BC = \sqrt{(3 - 2)^2 + (1 - (-3))^2 + (-2 - 1)^2} = \sqrt{1^2 + 4^2 + (-3)^2} = \sqrt{1 + 16 + 9} = \sqrt{26}$.
$3$. Length of $AC$:
$AC = \sqrt{(3 - (-1))^2 + (1 - 2)^2 + (-2 - 3)^2} = \sqrt{4^2 + (-1)^2 + (-5)^2} = \sqrt{16 + 1 + 25} = \sqrt{42}$.
Since all sides $AB = \sqrt{38}$,$BC = \sqrt{26}$,and $AC = \sqrt{42}$ are unequal,the triangle is a scalene triangle.

(D) Let the vertices be $A(2, -1, 1)$,$B(1, -3, -5)$,and $C(3, -4, -4)$.
First,calculate the side lengths:
$AB = \sqrt{(1-2)^2 + (-3-(-1))^2 + (-5-1)^2} = \sqrt{(-1)^2 + (-2)^2 + (-6)^2} = \sqrt{1 + 4 + 36} = \sqrt{41}$.
$BC = \sqrt{(3-1)^2 + (-4-(-3))^2 + (-4-(-5))^2} = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.
$AC = \sqrt{(3-2)^2 + (-4-(-1))^2 + (-4-1)^2} = \sqrt{1^2 + (-3)^2 + (-5)^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$.
Observe that $AB^2 = 41$ and $BC^2 + AC^2 = 6 + 35 = 41$.
Since $AB^2 = BC^2 + AC^2$,the triangle is a right-angled triangle with the right angle at $C$.
For a right-angled triangle,the circumradius $R$ is half of the hypotenuse.
Here,the hypotenuse is $AB = \sqrt{41}$.
Therefore,$R = \frac{AB}{2} = \frac{\sqrt{41}}{2}$.

(C) Let the vertices of $\triangle ABC$ be $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$.
Given that the midpoints of $AB, BC, CA$ are $(l, 0, 0), (0, m, 0), (0, 0, n)$ respectively.
For midpoint of $AB$: $\frac{x_1+x_2}{2}=l, \frac{y_1+y_2}{2}=0, \frac{z_1+z_2}{2}=0 \Rightarrow x_1+x_2=2l, y_1+y_2=0, z_1+z_2=0$ ... $(i)$
For midpoint of $BC$: $\frac{x_2+x_3}{2}=0, \frac{y_2+y_3}{2}=m, \frac{z_2+z_3}{2}=0 \Rightarrow x_2+x_3=0, y_2+y_3=2m, z_2+z_3=0$ ... (ii)
For midpoint of $CA$: $\frac{x_3+x_1}{2}=0, \frac{y_3+y_1}{2}=0, \frac{z_3+z_1}{2}=n \Rightarrow x_3+x_1=0, y_3+y_1=0, z_3+z_1=2n$ ... (iii)
Solving these equations:
Adding $(i)$,(ii),and (iii): $2(x_1+x_2+x_3)=2l \Rightarrow x_1+x_2+x_3=l$. Similarly,$y_1+y_2+y_3=m$ and $z_1+z_2+z_3=n$.
Subtracting (ii) from the sum: $x_1=l, y_1=-m, z_1=n$ (Wait,let's re-evaluate).
From $(i)$,$x_2=2l-x_1$. Substituting into (ii): $2l-x_1+x_3=0 \Rightarrow x_1-x_3=2l$. From (iii),$x_1+x_3=0$. Adding gives $2x_1=2l \Rightarrow x_1=l, x_3=-l, x_2=l$.
Similarly,$y_1=m, y_2=-m, y_3=m$ and $z_1=-n, z_2=n, z_3=n$.
Thus,$A(l, m, -n), B(l, -m, n), C(-l, m, n)$.
$AB^2 = (l-l)^2 + (m-(-m))^2 + (-n-n)^2 = 0 + 4m^2 + 4n^2 = 4(m^2+n^2)$.
$BC^2 = (l-(-l))^2 + (-m-m)^2 + (n-n)^2 = 4l^2 + 4m^2 + 0 = 4(l^2+m^2)$.
$CA^2 = (-l-l)^2 + (m-m)^2 + (n-(-n))^2 = 4l^2 + 0 + 4n^2 = 4(l^2+n^2)$.
$AB^2+BC^2+CA^2 = 4(m^2+n^2+l^2+m^2+l^2+n^2) = 8(l^2+m^2+n^2)$.
Therefore,$\frac{AB^2+BC^2+CA^2}{l^2+m^2+n^2} = \frac{8(l^2+m^2+n^2)}{l^2+m^2+n^2} = 8$.

(C) The coordinates of the centroid $G$ of a triangle with vertices $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$ are given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
Given $A(4, x, 1)$,$B(y, -5, 2)$,$C(7, 8, 3)$,and $G(3, 5, 2)$,we have:
$G = \left(\frac{4+y+7}{3}, \frac{x-5+8}{3}, \frac{1+2+3}{3}\right) = (3, 5, 2)$
Equating the coordinates:
$\frac{11+y}{3} = 3 \Rightarrow 11+y = 9 \Rightarrow y = -2$
$\frac{x+3}{3} = 5 \Rightarrow x+3 = 15 \Rightarrow x = 12$
Since $CG$ is a median,$F$ is the midpoint of $AB$. The coordinates of $F$ are:
$F = \left(\frac{4+y}{2}, \frac{x-5}{2}, \frac{1+2}{2}\right)$
Substituting $x=12$ and $y=-2$:
$F = \left(\frac{4-2}{2}, \frac{12-5}{2}, \frac{3}{2}\right) = \left(\frac{2}{2}, \frac{7}{2}, \frac{3}{2}\right) = \left(1, \frac{7}{2}, \frac{3}{2}\right)$.

(D) The centroid of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
For the given vertices $(4, p, -3)$,$(-1, -1, 2)$,and $(3, 5, -8)$,the centroid is $\left(\frac{4-1+3}{3}, \frac{p-1+5}{3}, \frac{-3+2-8}{3}\right) = \left(2, \frac{p+4}{3}, -3\right)$.
The mid-point of $(1, 4, -2)$ and $(q, 2, -4)$ is $\left(\frac{1+q}{2}, \frac{4+2}{2}, \frac{-2-4}{2}\right) = \left(\frac{1+q}{2}, 3, -3\right)$.
Equating the coordinates of the centroid and the mid-point:
$2 = \frac{1+q}{2} \Rightarrow 4 = 1+q \Rightarrow q = 3$.
$\frac{p+4}{3} = 3 \Rightarrow p+4 = 9 \Rightarrow p = 5$.
Thus,$p^2 + q^2 = 5^2 + 3^2 = 25 + 9 = 34$.

(A) Let the vertices of the triangle be $A(x, y, z)$,$B(-2, 3, 4)$,and $C(3, -1, 5)$.
The centroid $G$ of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
Given that the centroid is the origin $(0, 0, 0)$,we have:
$\frac{x-2+3}{3} = 0 \Rightarrow x+1 = 0 \Rightarrow x = -1$
$\frac{y+3-1}{3} = 0 \Rightarrow y+2 = 0 \Rightarrow y = -2$
$\frac{z+4+5}{3} = 0 \Rightarrow z+9 = 0 \Rightarrow z = -9$
Therefore,the third vertex is $(-1, -2, -9)$.

(C) The centroid $(x, y, z)$ of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by:
$x = \frac{x_1+x_2+x_3}{3}$,$y = \frac{y_1+y_2+y_3}{3}$,$z = \frac{z_1+z_2+z_3}{3}$
Since the origin $(0, 0, 0)$ is the centroid,we have:
$0 = \frac{2a - 4 + 8}{3} \Rightarrow 2a + 4 = 0 \Rightarrow a = -2$
$0 = \frac{2 + 3b + 14}{3} \Rightarrow 3b + 16 = 0 \Rightarrow b = -\frac{16}{3}$
$0 = \frac{6 - 10 + 2c}{3} \Rightarrow 2c - 4 = 0 \Rightarrow c = 2$
Thus,the values are $a = -2, b = -\frac{16}{3}, c = 2$.

(A) Let the points be $A(5, -4, 5)$,$B(-3, -3, 2)$,and $C(-1, -6, 8)$.
We calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.
$AB = \sqrt{(-3-5)^2 + (-3-(-4))^2 + (2-5)^2} = \sqrt{(-8)^2 + (1)^2 + (-3)^2} = \sqrt{64 + 1 + 9} = \sqrt{74}$.
$BC = \sqrt{(-1-(-3))^2 + (-6-(-3))^2 + (8-2)^2} = \sqrt{(2)^2 + (-3)^2 + (6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.
$CA = \sqrt{(5-(-1))^2 + (-4-(-6))^2 + (5-8)^2} = \sqrt{(6)^2 + (2)^2 + (-3)^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$.
Since $BC = CA = 7$,two sides of the triangle are equal.
Therefore,the points form an isosceles triangle.

(D) Let the vertices of the parallelogram be $A(2, 4, -1)$,$B(3, 6, -1)$,$C(4, 5, 1)$,and $D(x, y, z)$.
Since the diagonals of a parallelogram bisect each other,the midpoint of diagonal $AC$ is the same as the midpoint of diagonal $BD$.
Midpoint of $AC = \left( \frac{2+4}{2}, \frac{4+5}{2}, \frac{-1+1}{2} \right) = \left( 3, \frac{9}{2}, 0 \right)$.
Midpoint of $BD = \left( \frac{3+x}{2}, \frac{6+y}{2}, \frac{-1+z}{2} \right)$.
Equating the midpoints:
$\frac{3+x}{2} = 3 \implies 3+x = 6 \implies x = 3$.
$\frac{6+y}{2} = \frac{9}{2} \implies 6+y = 9 \implies y = 3$.
$\frac{-1+z}{2} = 0 \implies -1+z = 0 \implies z = 1$.
Thus,the fourth vertex $D$ is $(3, 3, 1)$.
Therefore,option $(D)$ is correct.

(D) Given points are $A(2, -1, 4)$,$B(1, 0, -1)$,$C(1, 2, 3)$,and $D(2, 1, 8)$.
First,we calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.
$AB = \sqrt{(1-2)^2 + (0-(-1))^2 + (-1-4)^2} = \sqrt{(-1)^2 + 1^2 + (-5)^2} = \sqrt{1+1+25} = \sqrt{27}$.
$BC = \sqrt{(1-1)^2 + (2-0)^2 + (3-(-1))^2} = \sqrt{0^2 + 2^2 + 4^2} = \sqrt{0+4+16} = \sqrt{20}$.
$CD = \sqrt{(2-1)^2 + (1-2)^2 + (8-3)^2} = \sqrt{1^2 + (-1)^2 + 5^2} = \sqrt{1+1+25} = \sqrt{27}$.
$DA = \sqrt{(2-2)^2 + (-1-1)^2 + (4-8)^2} = \sqrt{0^2 + (-2)^2 + (-4)^2} = \sqrt{0+4+16} = \sqrt{20}$.
Since $AB = CD = \sqrt{27}$ and $BC = DA = \sqrt{20}$,the opposite sides are equal.
Now,we check the diagonals $AC$ and $BD$.
$AC = \sqrt{(1-2)^2 + (2-(-1))^2 + (3-4)^2} = \sqrt{(-1)^2 + 3^2 + (-1)^2} = \sqrt{1+9+1} = \sqrt{11}$.
$BD = \sqrt{(2-1)^2 + (1-0)^2 + (8-(-1))^2} = \sqrt{1^2 + 1^2 + 9^2} = \sqrt{1+1+81} = \sqrt{83}$.
Since $AC \neq BD$,the diagonals are not equal.
Therefore,the points $A, B, C$,and $D$ form a parallelogram.

(A) Let $A = (1,2,3)$,$B = (3,-1,5)$,and $C = (4,0,-3)$.
First,we find the direction ratios of the sides:
Direction ratios of $\overline{AB} = (3-1, -1-2, 5-3) = (2, -3, 2)$.
Direction ratios of $\overline{AC} = (4-1, 0-2, -3-3) = (3, -2, -6)$.
Check for orthogonality: $\vec{AB} \cdot \vec{AC} = (2)(3) + (-3)(-2) + (2)(-6) = 6 + 6 - 12 = 0$.
Since the dot product is $0$,$\overline{AB} \perp \overline{AC}$,which means $\angle A = 90^{\circ}$.
In a right-angled triangle,the orthocentre $H$ is the vertex where the right angle is located. Thus,$H = A = (1, 2, 3)$.
The circumcentre $S$ of a right-angled triangle is the midpoint of the hypotenuse $\overline{BC}$.
$S = \left( \frac{3+4}{2}, \frac{-1+0}{2}, \frac{5-3}{2} \right) = \left( \frac{7}{2}, -\frac{1}{2}, 1 \right)$.
The distance $HS$ is given by the distance formula:
$HS = \sqrt{\left( \frac{7}{2} - 1 \right)^2 + \left( -\frac{1}{2} - 2 \right)^2 + (1 - 3)^2}$
$HS = \sqrt{\left( \frac{5}{2} \right)^2 + \left( -\frac{5}{2} \right)^2 + (-2)^2}$
$HS = \sqrt{\frac{25}{4} + \frac{25}{4} + 4} = \sqrt{\frac{50}{4} + \frac{16}{4}} = \sqrt{\frac{66}{4}} = \sqrt{\frac{33}{2}}$.
Thus,the correct option is $(A)$.

(D) Let the vertices of the triangle be $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$.
The mid-points of sides $AB, BC, CA$ are given as $D(1, 5, -1)$,$E(0, 4, -2)$,and $F(2, 3, 4)$ respectively.
By the mid-point formula:
For $AB$: $\frac{x_1+x_2}{2} = 1, \frac{y_1+y_2}{2} = 5, \frac{z_1+z_2}{2} = -1 \Rightarrow x_1+x_2 = 2, y_1+y_2 = 10, z_1+z_2 = -2$.
For $BC$: $\frac{x_2+x_3}{2} = 0, \frac{y_2+y_3}{2} = 4, \frac{z_2+z_3}{2} = -2 \Rightarrow x_2+x_3 = 0, y_2+y_3 = 8, z_2+z_3 = -4$.
For $CA$: $\frac{x_3+x_1}{2} = 2, \frac{y_3+y_1}{2} = 3, \frac{z_3+z_1}{2} = 4 \Rightarrow x_3+x_1 = 4, y_3+y_1 = 6, z_3+z_1 = 8$.
Adding these equations:
$2(x_1+x_2+x_3) = 2+0+4 = 6 \Rightarrow x_1+x_2+x_3 = 3$.
$2(y_1+y_2+y_3) = 10+8+6 = 24 \Rightarrow y_1+y_2+y_3 = 12$.
$2(z_1+z_2+z_3) = -2-4+8 = 2 \Rightarrow z_1+z_2+z_3 = 1$.
Now,to find $C(x_3, y_3, z_3)$,subtract the equations for $AB$ from the sums:
$x_3 = (x_1+x_2+x_3) - (x_1+x_2) = 3 - 2 = 1$.
$y_3 = (y_1+y_2+y_3) - (y_1+y_2) = 12 - 10 = 2$.
$z_3 = (z_1+z_2+z_3) - (z_1+z_2) = 1 - (-2) = 3$.
So,$C = (1, 2, 3)$.
The median from $C$ to $AB$ is the line segment $CD$,where $D$ is the mid-point of $AB$,$D(1, 5, -1)$.
Length $CD = \sqrt{(1-1)^2 + (5-2)^2 + (-1-3)^2} = \sqrt{0^2 + 3^2 + (-4)^2} = \sqrt{0 + 9 + 16} = \sqrt{25} = 5$.

(D) Let the points be $A(2, 3, 5)$,$B(-1, 5, -1)$,and $C(4, -3, 2)$.
Calculate the squared distances between the points using the distance formula $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$:
$AB^2 = (-1 - 2)^2 + (5 - 3)^2 + (-1 - 5)^2 = (-3)^2 + (2)^2 + (-6)^2 = 9 + 4 + 36 = 49$.
$BC^2 = (4 - (-1))^2 + (-3 - 5)^2 + (2 - (-1))^2 = (5)^2 + (-8)^2 + (3)^2 = 25 + 64 + 9 = 98$.
$AC^2 = (4 - 2)^2 + (-3 - 3)^2 + (2 - 5)^2 = (2)^2 + (-6)^2 + (-3)^2 = 4 + 36 + 9 = 49$.
Since $AB^2 = AC^2 = 49$,the triangle is isosceles because $AB = AC = 7$.
Check for the right-angled condition: $AB^2 + AC^2 = 49 + 49 = 98 = BC^2$.
Since the sum of the squares of two sides equals the square of the third side,the triangle is a right-angled triangle.
Thus,the triangle is an isosceles right-angled triangle.

(C) Let the circumcentre be $P(x, y, z)$. The circumcentre is equidistant from the vertices $A, B$,and $C$. Thus,$PA^2 = PB^2 = PC^2$.
$PA^2 = (x-3)^2 + (y-4)^2 + (z-5)^2$
$PB^2 = (x-2)^2 + (y-3)^2 + (z-1)^2$
$PC^2 = (x+1)^2 + (y-6)^2 + (z-1)^2$
Equating $PB^2 = PC^2$:
$(x-2)^2 + (y-3)^2 + (z-1)^2 = (x+1)^2 + (y-6)^2 + (z-1)^2$
$x^2 - 4x + 4 + y^2 - 6y + 9 = x^2 + 2x + 1 + y^2 - 12y + 36$
$-6x + 6y = 24 \implies -x + y = 4 \implies y = x + 4$.
Equating $PA^2 = PB^2$:
$(x-3)^2 + (y-4)^2 + (z-5)^2 = (x-2)^2 + (y-3)^2 + (z-1)^2$
Substitute $y = x+4$:
$(x-3)^2 + (x+4-4)^2 + (z-5)^2 = (x-2)^2 + (x+4-3)^2 + (z-1)^2$
$(x-3)^2 + x^2 + (z-5)^2 = (x-2)^2 + (x+1)^2 + (z-1)^2$
$x^2 - 6x + 9 + x^2 + z^2 - 10z + 25 = x^2 - 4x + 4 + x^2 + 2x + 1 + z^2 - 2z + 1$
$-6x - 10z + 34 = -2x - 2z + 6$
$-4x - 8z = -28 \implies x + 2z = 7 \implies z = \frac{7-x}{2}$.
Since $P$ lies on the plane of the triangle,we check the options. For option $C(1, 5, 3)$:
$y = 1+4 = 5$ (Matches).
$z = (7-1)/2 = 3$ (Matches).
Thus,the circumcentre is $(1, 5, 3)$.

(B) Let $A \equiv (x_1, y_1, z_1)$,$B \equiv (x_2, y_2, z_2)$,and $C \equiv (x_3, y_3, z_3)$.
Since $F(0, 1, 0)$ is the mid-point of $AB$,we have:
$x_1 + x_2 = 0, y_1 + y_2 = 2, z_1 + z_2 = 0$ $(1)$
Since $D(2, 1, 0)$ is the mid-point of $BC$,we have:
$x_2 + x_3 = 4, y_2 + y_3 = 2, z_2 + z_3 = 0$ $(2)$
Since $E(2, 0, 0)$ is the mid-point of $AC$,we have:
$x_3 + x_1 = 4, y_3 + y_1 = 0, z_3 + z_1 = 0$ $(3)$
Adding $(1)$,$(2)$,and $(3)$,we get:
$2(x_1 + x_2 + x_3) = 8 \implies x_1 + x_2 + x_3 = 4$
$2(y_1 + y_2 + y_3) = 4 \implies y_1 + y_2 + y_3 = 2$
$2(z_1 + z_2 + z_3) = 0 \implies z_1 + z_2 + z_3 = 0$
The centroid of $\triangle ABC$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
Substituting the sums,the centroid is $\left(\frac{4}{3}, \frac{2}{3}, 0\right)$.

(C) Let the vertices of the triangle be $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$.
Given the midpoints of $AB, BC, CA$ are $(l, 0, 0), (0, m, 0), (0, 0, n)$ respectively.
Using the midpoint formula:
$x_1+x_2=2l, y_1+y_2=0, z_1+z_2=0$
$x_2+x_3=0, y_2+y_3=2m, z_2+z_3=0$
$x_1+x_3=0, y_1+y_3=0, z_1+z_3=2n$
Solving these systems of equations:
For $x$: $x_1+x_2=2l, x_2+x_3=0, x_1+x_3=0 \implies x_1=l, x_2=l, x_3=-l$
For $y$: $y_1+y_2=0, y_2+y_3=2m, y_1+y_3=0 \implies y_1=-m, y_2=m, y_3=m$
For $z$: $z_1+z_2=0, z_2+z_3=0, z_1+z_3=2n \implies z_1=n, z_2=-n, z_3=n$
Thus,the vertices are $A(l, -m, n), B(l, m, -n), C(-l, m, n)$.
Calculating the squared side lengths:
$AB^2 = (l-l)^2 + (m-(-m))^2 + (-n-n)^2 = 0 + (2m)^2 + (-2n)^2 = 4m^2 + 4n^2$
$BC^2 = (-l-l)^2 + (m-m)^2 + (n-(-n))^2 = (-2l)^2 + 0 + (2n)^2 = 4l^2 + 4n^2$
$CA^2 = (l-(-l))^2 + (-m-m)^2 + (n-n)^2 = (2l)^2 + (-2m)^2 + 0 = 4l^2 + 4m^2$
Summing these:
$AB^2+BC^2+CA^2 = (4m^2+4n^2) + (4l^2+4n^2) + (4l^2+4m^2) = 8(l^2+m^2+n^2)$
Therefore,$\frac{AB^2+BC^2+CA^2}{l^2+m^2+n^2} = \frac{8(l^2+m^2+n^2)}{l^2+m^2+n^2} = 8$.

(A) Let $D$ be the midpoint of $BC$. The coordinates of $D$ are $\left(\frac{\alpha+7}{2}, \frac{3+5}{2}, \frac{3+\beta}{2}\right) = \left(\frac{\alpha+7}{2}, 4, \frac{3+\beta}{2}\right)$.
The direction ratios of the median $AD$ are given by the difference of coordinates of $D$ and $A$:
$\left(\frac{\alpha+7}{2} - 2, 4 - 3, \frac{3+\beta}{2} - 5\right) = \left(\frac{\alpha+3}{2}, 1, \frac{\beta-7}{2}\right)$.
Since the median $AD$ is equally inclined to the coordinate axes,its direction cosines are equal,which implies that its direction ratios must be proportional to $(1, 1, 1)$.
Thus,$\frac{\alpha+3}{2} = 1$ and $\frac{\beta-7}{2} = 1$.
Solving for $\alpha$: $\alpha+3 = 2 \Rightarrow \alpha = -1$.
Solving for $\beta$: $\beta-7 = 2 \Rightarrow \beta = 9$.
Therefore,$\frac{\beta}{\alpha} = \frac{9}{-1} = -9$.

(D) Let $A=(2,3,-4), B=(-3,3,-2), C=(-1,4,2), D=(3,5,1)$.
$G_1$ is the centroid of face $ABD$: $G_1 = \left(\frac{2-3+3}{3}, \frac{3+3+5}{3}, \frac{-4-2+1}{3}\right) = \left(\frac{2}{3}, \frac{11}{3}, -\frac{5}{3}\right)$.
$G_2$ is the centroid of face $BCD$: $G_2 = \left(\frac{-3-1+3}{3}, \frac{3+4+5}{3}, \frac{-2+2+1}{3}\right) = \left(-\frac{1}{3}, 4, \frac{1}{3}\right)$.
$G_3$ is the centroid of face $ACD$: $G_3 = \left(\frac{2-1+3}{3}, \frac{3+4+5}{3}, \frac{-4+2+1}{3}\right) = \left(\frac{4}{3}, 4, -\frac{1}{3}\right)$.
Let $G$ be the centroid of $\Delta G_1 G_2 G_3$:
$G = \left(\frac{\frac{2}{3} - \frac{1}{3} + \frac{4}{3}}{3}, \frac{\frac{11}{3} + 4 + 4}{3}, \frac{-\frac{5}{3} + \frac{1}{3} - \frac{1}{3}}{3}\right)$
$G = \left(\frac{\frac{5}{3}}{3}, \frac{\frac{11+24}{3}}{3}, \frac{-\frac{5}{3}}{3}\right) = \left(\frac{5}{9}, \frac{35}{9}, -\frac{5}{9}\right)$.

(D) Let the vertices of the tetrahedron be $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,$C(x_3, y_3, z_3)$,and $D(x_4, y_4, z_4)$.
Since the coordinates of each vertex are in arithmetic progression,we have $y_i = x_i + d_i$ and $z_i = x_i + 2d_i$ for $i = 1, 2, 3, 4$.
However,the problem implies a common difference $d$ for the coordinates of each vertex. Thus,$y_i = x_i + d$ and $z_i = x_i + 2d$.
The centroid $G$ is given by $\left(\frac{\sum x_i}{4}, \frac{\sum y_i}{4}, \frac{\sum z_i}{4}\right) = (2, 3, k)$.
From the $x$-coordinate: $\frac{\sum x_i}{4} = 2 \implies \sum x_i = 8$.
From the $y$-coordinate: $\frac{\sum (x_i + d)}{4} = 3 \implies \frac{\sum x_i + 4d}{4} = 3 \implies \frac{8 + 4d}{4} = 3 \implies 2 + d = 3 \implies d = 1$.
From the $z$-coordinate: $k = \frac{\sum (x_i + 2d)}{4} = \frac{\sum x_i + 8d}{4} = \frac{8 + 8(1)}{4} = \frac{16}{4} = 4$.
Thus,the centroid $G$ is $(2, 3, 4)$.
The distance of $G$ from the origin $(0, 0, 0)$ is $\sqrt{2^2 + 3^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$.

(A) Given that $ABCD$ is a parallelogram and $E$ is the mid-point of $AB$.
Since $AB \parallel CD$,we have $AE \parallel CD$.
In $\triangle PAE$ and $\triangle PCD$,
$\angle PAE = \angle PCD$ (alternate interior angles)
$\angle APE = \angle CPD$ (vertically opposite angles)
Therefore,$\triangle PAE \sim \triangle PCD$ by $AA$ similarity.
From the property of similar triangles,we have:
$\frac{PA}{PC} = \frac{PE}{PD} = \frac{AE}{CD}$.
Since $E$ is the mid-point of $AB$,$AE = \frac{1}{2} AB$. Also,$AB = CD$ in a parallelogram,so $AE = \frac{1}{2} CD$.
Thus,$\frac{PA}{PC} = \frac{PE}{PD} = \frac{1}{2}$.
This implies $\frac{PA}{PC} = \frac{1}{2}$ and $\frac{PD}{PE} = 2$.
Therefore,$\frac{DP}{PE} + \frac{AP}{PC} = 2 + \frac{1}{2} = \frac{5}{2}$.

(C) Given,$ABCD$ is a parallelogram with vertices $A(4,4,-1)$,$B(5,6,-1)$,$C(6,5,1)$ and $D(x, y, z)$.
We know that the diagonals of a parallelogram $ABCD$ bisect each other.
Therefore,the mid-point of $AC$ = mid-point of $BD$.
$\left(\frac{4+6}{2}, \frac{4+5}{2}, \frac{-1+1}{2}\right) = \left(\frac{x+5}{2}, \frac{y+6}{2}, \frac{z-1}{2}\right)$
$\left(\frac{10}{2}, \frac{9}{2}, 0\right) = \left(\frac{x+5}{2}, \frac{y+6}{2}, \frac{z-1}{2}\right)$
On comparing both sides,we get:
$\frac{x+5}{2} = \frac{10}{2}$ $\Rightarrow x+5 = 10$ $\Rightarrow x = 5$
$\frac{y+6}{2} = \frac{9}{2}$ $\Rightarrow y+6 = 9$ $\Rightarrow y = 3$
$\frac{z-1}{2} = 0$ $\Rightarrow z-1 = 0$ $\Rightarrow z = 1$
Thus,the vertex $D(x, y, z)$ is $(5, 3, 1)$.

(B) First,calculate the lengths of the sides of $\triangle ABC$:
$c = AB = \sqrt{(2-1)^2 + (3-2)^2 + (-1 - (-3))^2} = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6}$
$a = BC = \sqrt{(3-2)^2 + (1-3)^2 + (1 - (-1))^2} = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{9} = 3$
$b = AC = \sqrt{(3-1)^2 + (1-2)^2 + (1 - (-3))^2} = \sqrt{2^2 + (-1)^2 + 4^2} = \sqrt{21}$
Using the cosine rule:
$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{21 + 6 - 9}{2 \times \sqrt{21} \times \sqrt{6}} = \frac{18}{2 \sqrt{126}} = \frac{9}{3 \sqrt{14}} = \frac{3}{\sqrt{14}}$
$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{9 + 6 - 21}{2 \times 3 \times \sqrt{6}} = \frac{-6}{6 \sqrt{6}} = -\frac{1}{\sqrt{6}}$
Therefore,$\left|\frac{\cos A}{\cos B}\right| = \left|\frac{3}{\sqrt{14}} \div \left(-\frac{1}{\sqrt{6}}\right)\right| = \left| -\frac{3 \sqrt{6}}{\sqrt{14}} \right| = \frac{3 \sqrt{3 \times 2}}{\sqrt{7 \times 2}} = \frac{3 \sqrt{3}}{\sqrt{7}}$.

(B) Let the given points be $A(2,-1,3), B(4,-2,1), C(4,5,-7)$ and $D(2,6,-5)$.
First,we find the mid-points of the diagonals $AC$ and $BD$:
Mid-point of $AC = \left(\frac{2+4}{2}, \frac{-1+5}{2}, \frac{3-7}{2}\right) = (3, 2, -2)$.
Mid-point of $BD = \left(\frac{4+2}{2}, \frac{-2+6}{2}, \frac{1-5}{2}\right) = (3, 2, -2)$.
Since the mid-points of the diagonals are the same,the quadrilateral $ABCD$ is a parallelogram.
Next,we check the side lengths:
$AB = \sqrt{(4-2)^2 + (-2 - (-1))^2 + (1-3)^2} = \sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{4+1+4} = \sqrt{9} = 3$.
$BC = \sqrt{(4-4)^2 + (5 - (-2))^2 + (-7-1)^2} = \sqrt{0^2 + 7^2 + (-8)^2} = \sqrt{0+49+64} = \sqrt{113}$.
Since $AB \neq BC$,it is not a rhombus or a square.
Now,check if it is a rectangle by calculating the dot product of adjacent sides:
$\vec{AB} = (4-2)\hat{i} + (-2+1)\hat{j} + (1-3)\hat{k} = 2\hat{i} - \hat{j} - 2\hat{k}$.
$\vec{BC} = (4-4)\hat{i} + (5+2)\hat{j} + (-7-1)\hat{k} = 0\hat{i} + 7\hat{j} - 8\hat{k}$.
$\vec{AB} \cdot \vec{BC} = (2)(0) + (-1)(7) + (-2)(-8) = 0 - 7 + 16 = 9 \neq 0$.
Since the dot product is not zero,the sides are not perpendicular.
Therefore,$ABCD$ is a parallelogram.

(B) Given that $AB = AC$.
Using the distance formula,$AB^2 = AC^2$.
$AB^2 = (2 - (-1))^2 + (3 - k)^2 + (k - (-1))^2 = 3^2 + (3 - k)^2 + (k + 1)^2 = 9 + 9 - 6k + k^2 + k^2 + 2k + 1 = 2k^2 - 4k + 19$.
$AC^2 = (2 - 4)^2 + (3 - (-3))^2 + (k - 2)^2 = (-2)^2 + 6^2 + (k - 2)^2 = 4 + 36 + k^2 - 4k + 4 = k^2 - 4k + 44$.
Equating $AB^2 = AC^2$:
$2k^2 - 4k + 19 = k^2 - 4k + 44$.
$k^2 = 25$.
Since $k > 0$,we have $k = 5$.
Now,calculate the side lengths:
$AB^2 = 2(5)^2 - 4(5) + 19 = 50 - 20 + 19 = 49 \Rightarrow AB = 7$.
$AC^2 = 49 \Rightarrow AC = 7$.
$BC^2 = (-1 - 4)^2 + (5 - (-3))^2 + (-1 - 2)^2 = (-5)^2 + 8^2 + (-3)^2 = 25 + 64 + 9 = 98$.
Since $AB^2 + AC^2 = 49 + 49 = 98 = BC^2$,the triangle satisfies the Pythagorean theorem.
Thus,$\triangle ABC$ is a right-angled isosceles triangle.

(D) Let the vertices of the triangle be $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$.
Given the midpoints of $AB, BC, CA$ are $M_1(3,0,0)$,$M_2(0,4,0)$,and $M_3(0,0,5)$ respectively.
Using the midpoint formula:
$x_1+x_2=6, x_2+x_3=0, x_3+x_1=0$
Solving these,we get $x_1=3, x_2=3, x_3=-3$.
Similarly for $y$ coordinates:
$y_1+y_2=0, y_2+y_3=8, y_3+y_1=0$
Solving these,we get $y_1=-4, y_2=4, y_3=4$.
Similarly for $z$ coordinates:
$z_1+z_2=0, z_2+z_3=0, z_3+z_1=10$
Solving these,we get $z_1=5, z_2=-5, z_3=5$.
Thus,the vertices are $A(3, -4, 5)$,$B(3, 4, -5)$,and $C(-3, 4, 5)$.
Now,calculate the squares of the side lengths:
$AB^2 = (3-3)^2 + (4-(-4))^2 + (-5-5)^2 = 0^2 + 8^2 + (-10)^2 = 64 + 100 = 164$.
$BC^2 = (-3-3)^2 + (4-4)^2 + (5-(-5))^2 = (-6)^2 + 0^2 + 10^2 = 36 + 100 = 136$.
$CA^2 = (3-(-3))^2 + (-4-4)^2 + (5-5)^2 = 6^2 + (-8)^2 + 0^2 = 36 + 64 = 100$.
Finally,$AB^2+BC^2+CA^2 = 164 + 136 + 100 = 400$.