Triangle and Parallelogram Questions in English

(A) Let the vertices of the parallelogram be $A(x, -1)$,$B(3, y)$,$C(-2, 3)$,and $D(-3, -2)$.
In a parallelogram,the diagonals bisect each other,meaning the mid-points of the diagonals coincide.
The mid-point of diagonal $AC$ is $\left( \frac{x - 2}{2}, \frac{-1 + 3}{2} \right) = \left( \frac{x - 2}{2}, 1 \right)$.
The mid-point of diagonal $BD$ is $\left( \frac{3 - 3}{2}, \frac{y - 2}{2} \right) = \left( 0, \frac{y - 2}{2} \right)$.
Equating the mid-points: $\frac{x - 2}{2} = 0 \Rightarrow x = 2$.
And $1 = \frac{y - 2}{2}$ $\Rightarrow y - 2 = 2$ $\Rightarrow y = 4$.
Thus,$x = 2$ and $y = 4$.

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

(D) In a parallelogram $ABCD$,the diagonals $AC$ and $BD$ bisect each other at the same midpoint $M$.
The midpoint $M$ of diagonal $AC$ is given by:
$M = \left( \frac{3+7}{2}, \frac{5+10}{2} \right) = \left( 5, \frac{15}{2} \right)$
Let the fourth vertex be $D(x, y)$. Since $M$ is also the midpoint of diagonal $BD$:
$\frac{-5+x}{2} = 5 \implies -5+x = 10 \implies x = 15$
$\frac{-4+y}{2} = \frac{15}{2} \implies -4+y = 15 \implies y = 19$
Thus,the coordinates of the fourth vertex $D$ are $(15, 19)$.

(B) Let the vertices be $A(0, 1, 2)$,$B(2, -1, 3)$,and $C(1, -3, 1)$.
Calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.
$AB = \sqrt{(2-0)^2 + (-1-1)^2 + (3-2)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
$BC = \sqrt{(1-2)^2 + (-3 - (-1))^2 + (1-3)^2} = \sqrt{(-1)^2 + (-2)^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$.
$AC = \sqrt{(1-0)^2 + (-3-1)^2 + (1-2)^2} = \sqrt{1^2 + (-4)^2 + (-1)^2} = \sqrt{1 + 16 + 1} = \sqrt{18} = 3\sqrt{2}$.
Since $AB = BC = 3$,the triangle is isosceles.
Check for the right-angled condition: $AB^2 + BC^2 = 3^2 + 3^2 = 9 + 9 = 18$,and $AC^2 = (3\sqrt{2})^2 = 18$.
Since $AB^2 + BC^2 = AC^2$,the triangle is a right-angled triangle.
Therefore,the triangle is an isosceles right-angled triangle.

(D) Let the points be $A(-1, 3, 2)$,$B(-4, 2, -2)$,and $C(5, 5, \lambda)$.
Since the points are collinear,the direction ratios of line segment $AB$ must be proportional to the direction ratios of line segment $BC$.
The direction ratios of $AB$ are $(-4 - (-1), 2 - 3, -2 - 2) = (-3, -1, -4)$.
The direction ratios of $BC$ are $(5 - (-4), 5 - 2, \lambda - (-2)) = (9, 3, \lambda + 2)$.
For collinearity,the ratio of corresponding direction ratios must be equal:
$\frac{-3}{9} = \frac{-1}{3} = \frac{-4}{\lambda + 2}$.
From $\frac{-1}{3} = \frac{-4}{\lambda + 2}$,we get:
$-1(\lambda + 2) = -4 \times 3$
$-\lambda - 2 = -12$
$-\lambda = -10$
$\lambda = 10$.

(C) 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 the formula $G = (\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3})$.
Given that the centroid is the origin $(0, 0, 0)$,we have:
$\frac{a - 2 + 4}{3} = 0 \Rightarrow a + 2 = 0 \Rightarrow a = -2$.
$\frac{1 + b + 7}{3} = 0 \Rightarrow b + 8 = 0 \Rightarrow b = -8$.
$\frac{3 - 5 + c}{3} = 0 \Rightarrow c - 2 = 0 \Rightarrow c = 2$.
Thus,the values are $a = -2, b = -8, c = 2$.

(C) Points $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$ are collinear if the direction ratios of $AB$ are proportional to the direction ratios of $BC$.
For option $(A)$: Points $(1, -1, 1), (-1, 1, 1), (0, 0, 1)$. Direction ratios of $AB$ are $(-2, 2, 0)$ and $BC$ are $(1, -1, 0)$. Since $(-2)/1 = 2/(-1) = 0/0$ (undefined),these are collinear.
For option $(B)$: Points $(1, 2, 3), (3, 2, 1), (2, 2, 2)$. Direction ratios of $AB$ are $(2, 0, -2)$ and $BC$ are $(-1, 0, 1)$. Since $2/(-1) = 0/0 = -2/1$,these are collinear.
For option $(C)$: Points $A(-2, 4, -3), B(4, -3, -2), C(-3, -2, 4)$. Direction ratios of $AB$ are $(6, -7, 1)$ and $BC$ are $(-7, 1, 6)$. Since $6/(-7) \neq -7/1 \neq 1/6$,these points are non-collinear.
For option $(D)$: Points $(2, 0, -1), (3, 2, -2), (5, 6, -4)$. Direction ratios of $AB$ are $(1, 2, -1)$ and $BC$ are $(2, 4, -2)$. Since $1/2 = 2/4 = -1/(-2)$,these are collinear.

(A) Let the points be $A(-2, 4, 7)$,$B(3, -6, -8)$,and $C(1, -2, -2)$.
To check if the points are collinear,we calculate the direction ratios of lines $AB$ and $BC$.
The direction ratios of line $AB$ are $(3 - (-2), -6 - 4, -8 - 7) = (5, -10, -15)$.
The direction ratios of line $BC$ are $(1 - 3, -2 - (-6), -2 - (-8)) = (-2, 4, 6)$.
We observe that the ratio of the direction ratios is $\frac{5}{-2} = \frac{-10}{4} = \frac{-15}{6} = -2.5$.
Since the direction ratios are proportional,the lines $AB$ and $BC$ are parallel.
Because they share a common point $B$,the points $A, B$,and $C$ must be collinear.

(B) Let the vertices be $A(1, 1, 1)$,$B(-2, 4, 1)$,$C(-1, 5, 5)$,and $D(2, 2, 5)$.
Calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$:
$AB = \sqrt{(-2-1)^2 + (4-1)^2 + (1-1)^2} = \sqrt{(-3)^2 + 3^2 + 0^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$.
$BC = \sqrt{(-1-(-2))^2 + (5-4)^2 + (5-1)^2} = \sqrt{1^2 + 1^2 + 4^2} = \sqrt{1+1+16} = \sqrt{18} = 3\sqrt{2}$.
$CD = \sqrt{(2-(-1))^2 + (2-5)^2 + (5-5)^2} = \sqrt{3^2 + (-3)^2 + 0^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$.
$DA = \sqrt{(1-2)^2 + (1-2)^2 + (1-5)^2} = \sqrt{(-1)^2 + (-1)^2 + (-4)^2} = \sqrt{1+1+16} = \sqrt{18} = 3\sqrt{2}$.
Since all sides are equal,we check the diagonals $AC$ and $BD$:
$AC = \sqrt{(-1-1)^2 + (5-1)^2 + (5-1)^2} = \sqrt{(-2)^2 + 4^2 + 4^2} = \sqrt{4+16+16} = \sqrt{36} = 6$.
$BD = \sqrt{(2-(-2))^2 + (2-4)^2 + (5-1)^2} = \sqrt{4^2 + (-2)^2 + 4^2} = \sqrt{16+4+16} = \sqrt{36} = 6$.
Since all sides are equal $(3\sqrt{2})$ and both diagonals are equal $(6)$,the figure is a square.

(D) Let the vertices of the triangle be $A(0, 7, 10)$,$B(-1, 6, 6)$,and $C(-4, 9, 6)$.
Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$:
$AB = \sqrt{(-1-0)^2 + (6-7)^2 + (6-10)^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}$
$BC = \sqrt{(-4 - (-1))^2 + (9-6)^2 + (6-6)^2} = \sqrt{(-3)^2 + 3^2 + 0^2} = \sqrt{9 + 9 + 0} = \sqrt{18} = 3\sqrt{2}$
$AC = \sqrt{(-4-0)^2 + (9-7)^2 + (6-10)^2} = \sqrt{(-4)^2 + 2^2 + (-4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$
Now,check for the right-angled triangle condition using Pythagoras theorem:
$AB^2 + BC^2 = (3\sqrt{2})^2 + (3\sqrt{2})^2 = 18 + 18 = 36$
$AC^2 = 6^2 = 36$
Since $AB^2 + BC^2 = AC^2$,the triangle is right-angled.
Also,since $AB = BC = 3\sqrt{2}$,the triangle is isosceles.
Therefore,the triangle is a right-angled isosceles triangle.

(D) Let the given points be $A(5, -4, 2)$,$B(4, -3, 1)$,$C(7, -6, 4)$,and $D(8, -7, 5)$.
Calculate the vectors:
$\vec{AB} = (4-5, -3-(-4), 1-2) = (-1, 1, -1)$
$\vec{BC} = (7-4, -6-(-3), 4-1) = (3, -3, 3)$
$\vec{CD} = (8-7, -7-(-6), 5-4) = (1, -1, 1)$
$\vec{DA} = (5-8, -4-(-7), 2-5) = (-3, 3, -3)$
Note that $\vec{BC} = -3\vec{AB}$ and $\vec{DA} = 3\vec{AB}$.
Since the vectors are collinear,the points $A, B, C, D$ lie on a single straight line.
Therefore,they do not form a polygon (rectangle,square,or parallelogram).
Hence,option $(d)$ is correct.

(D) The direction ratios of line $AB$ are $(2 - (-1), 3 - 3, 5 - 2) = (3, 0, 3)$.
The direction ratios of line $AC$ are $(3 - (-1), 5 - 3, -2 - 2) = (4, 2, -4)$.
Let $\theta$ be the angle between $AB$ and $AC$. The formula for the cosine of the angle between two lines with direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$ is $\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}$.
Substituting the values: $\cos A = \frac{|(3)(4) + (0)(2) + (3)(-4)|}{\sqrt{3^2 + 0^2 + 3^2} \sqrt{4^2 + 2^2 + (-4)^2}}$.
$\cos A = \frac{|12 + 0 - 12|}{\sqrt{18} \sqrt{16 + 4 + 16}} = \frac{0}{\sqrt{18} \sqrt{36}} = 0$.
Since $\cos A = 0$,we have $A = 90^o$.

(C) Let the coordinates of the point be $P(x, y, z)$.
The distance of point $P(x, y, z)$ from the $xy$-plane is $|z|$.
The distance of point $P(x, y, z)$ from the $yz$-plane is $|x|$.
The distance of point $P(x, y, z)$ from the $zx$-plane is $|y|$.
According to the problem,the sum of its distances from the $xy$-plane and $yz$-plane is equal to its distance from the $zx$-plane:
$|z| + |x| = |y|$.
Assuming the point lies in the first octant where $x, y, z > 0$,we have $z + x = y$.
Rearranging the terms,we get $x - y + z = 0$.

(B) Let the vertices of the parallelogram $ABCD$ be $A(-1, 0), B(3, 1), C(2, 2),$ and $D(x, y).$
Since the diagonals of a parallelogram bisect each other,the midpoint of $AC$ is the same as the midpoint of $BD.$
Midpoint of $AC = (\frac{-1+2}{2}, \frac{0+2}{2}) = (\frac{1}{2}, 1).$
Midpoint of $BD = (\frac{3+x}{2}, \frac{1+y}{2}).$
Equating the midpoints: $(\frac{3+x}{2}, \frac{1+y}{2}) = (\frac{1}{2}, 1).$
For the $x$-coordinate: $\frac{3+x}{2} = \frac{1}{2}$ $\Rightarrow 3+x = 1$ $\Rightarrow x = -2.$
For the $y$-coordinate: $\frac{1+y}{2} = 1$ $\Rightarrow 1+y = 2$ $\Rightarrow y = 1.$
Thus,the fourth vertex is $(-2, 1).$

(A) For option $A$: The midpoint of diagonal $AC$ is $(\frac{-2+4}{2}, \frac{-1+3}{2}) = (1, 1)$. The midpoint of diagonal $BD$ is $(\frac{1+1}{2}, \frac{0+2}{2}) = (1, 1)$. Since the midpoints of the diagonals coincide,the diagonals bisect each other,so $ABCD$ is a parallelogram.
For option $B$: Calculating side lengths,$AB = \sqrt{(-2 - (-4))^2 + (-4 - (-1))^2} = \sqrt{2^2 + (-3)^2} = \sqrt{13}$. $BC = \sqrt{(4 - (-2))^2 + (0 - (-4))^2} = \sqrt{6^2 + 4^2} = \sqrt{52}$. Since $AB \neq BC$,it is not a square.
For option $C$: Calculating side lengths,$AB = \sqrt{(2-0)^2 + (1 - (-1))^2} = \sqrt{2^2 + 2^2} = \sqrt{8}$. $BC = \sqrt{(0-2)^2 + (3-1)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{8}$. $CD = \sqrt{(-2-0)^2 + (1-3)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{8}$. $DA = \sqrt{(0 - (-2))^2 + (-1-1)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{8}$. All sides are equal. The diagonal $AC = \sqrt{(0-0)^2 + (3 - (-1))^2} = 4$ and $BD = \sqrt{(-2-2)^2 + (1-1)^2} = 4$. Since diagonals are equal and sides are equal,it is a square,not a rhombus.
Thus,only statement $A$ is true.

(B) Let the vertices of the parallelogram be $A(1, 3)$,$B(2, 0)$,$C(5, 1)$,and $D(x, y)$.
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 $AC = (\frac{1+5}{2}, \frac{3+1}{2}) = (3, 2)$.
Midpoint of $BD = (\frac{2+x}{2}, \frac{0+y}{2})$.
Equating the midpoints:
$\frac{2+x}{2} = 3$ $\Rightarrow 2+x = 6$ $\Rightarrow x = 4$.
$\frac{0+y}{2} = 2 \Rightarrow y = 4$.
Thus,the fourth vertex is $(4, 4)$.

(C) Let $D$ be the mid-point 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:
$AD = \left( \frac{\lambda - 1}{2} - 2, 4 - 3, \frac{\mu + 2}{2} - 5 \right) = \left( \frac{\lambda - 5}{2}, 1, \frac{\mu - 8}{2} \right)$
Since the median $AD$ is equally inclined to the coordinate axes,its direction ratios must be equal in magnitude:
$\left| \frac{\lambda - 5}{2} \right| = |1| = \left| \frac{\mu - 8}{2} \right|$
This implies:
$\frac{\lambda - 5}{2} = \pm 1 \Rightarrow \lambda - 5 = \pm 2 \Rightarrow \lambda = 7 \text{ or } 3$
$\frac{\mu - 8}{2} = \pm 1 \Rightarrow \mu - 8 = \pm 2 \Rightarrow \mu = 10 \text{ or } 6$
For the median to be equally inclined,the direction ratios must be equal,i.e.,$\frac{\lambda - 5}{2} = 1 = \frac{\mu - 8}{2}$.
Thus,$\lambda = 7$ and $\mu = 10$.
Checking the options,for $\lambda = 7$ and $\mu = 10$:
$10\lambda - 7\mu = 10(7) - 7(10) = 70 - 70 = 0$.
Therefore,the correct relation is $10\lambda - 7\mu = 0$.

(B) Using the distance formula $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$,we calculate the squares of the sides:
$AB^2 = (10 - 3)^2 + (20 - 6)^2 + (30 - 9)^2 = 7^2 + 14^2 + 21^2 = 49 + 196 + 441 = 686$
$BC^2 = (25 - 10)^2 + (-41 - 20)^2 + (5 - 30)^2 = 15^2 + (-61)^2 + (-25)^2 = 225 + 3721 + 625 = 4571$
$CA^2 = (3 - 25)^2 + (6 - (-41))^2 + (9 - 5)^2 = (-22)^2 + 47^2 + 4^2 = 484 + 2209 + 16 = 2709$
For a right-angled triangle,the sum of the squares of two sides must equal the square of the third side (Pythagoras theorem).
Checking the sums:
$AB^2 + CA^2 = 686 + 2709 = 3395 \neq 4571 (BC^2)$
$AB^2 + BC^2 = 686 + 4571 = 5257 \neq 2709 (CA^2)$
$BC^2 + CA^2 = 4571 + 2709 = 7280 \neq 686 (AB^2)$
Since none of the combinations satisfy the condition $a^2 + b^2 = c^2$,the points $A, B, C$ do not form a right-angled triangle.

Let the points be $A(0, 7, -10)$,$B(1, 6, -6)$,and $C(4, 9, -6)$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$:
$AB = \sqrt{(1 - 0)^2 + (6 - 7)^2 + (-6 - (-10))^2} = \sqrt{1^2 + (-1)^2 + 4^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}$.
$BC = \sqrt{(4 - 1)^2 + (9 - 6)^2 + (-6 - (-6))^2} = \sqrt{3^2 + 3^2 + 0^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$.
$CA = \sqrt{(0 - 4)^2 + (7 - 9)^2 + (-10 - (-6))^2} = \sqrt{(-4)^2 + (-2)^2 + (-4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$.
Since $AB = BC = 3\sqrt{2}$ and $AB \neq CA$,two sides of the triangle are equal.
Therefore,the given points are the vertices of an isosceles triangle.

(N/A) Let the points be $A(0, 7, 10)$,$B(-1, 6, 6)$,and $C(-4, 9, 6)$.
Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$:
$AB = \sqrt{(-1-0)^2 + (6-7)^2 + (6-10)^2} = \sqrt{(-1)^2 + (-1)^2 + (-4)^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}$
$BC = \sqrt{(-4 - (-1))^2 + (9-6)^2 + (6-6)^2} = \sqrt{(-3)^2 + (3)^2 + (0)^2} = \sqrt{9 + 9 + 0} = \sqrt{18} = 3\sqrt{2}$
$CA = \sqrt{(0 - (-4))^2 + (7-9)^2 + (10-6)^2} = \sqrt{(4)^2 + (-2)^2 + (4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$
Now,check for the Pythagoras theorem condition $AB^2 + BC^2 = AC^2$:
$AB^2 + BC^2 = (3\sqrt{2})^2 + (3\sqrt{2})^2 = 18 + 18 = 36$
$AC^2 = (6)^2 = 36$
Since $AB^2 + BC^2 = AC^2$,the points $A, B,$ and $C$ form a right-angled triangle.

(N/A) Let the given points be $A(-1, 2, 1)$,$B(1, -2, 5)$,$C(4, -7, 8)$,and $D(2, -3, 4)$.
First,we calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$:
$AB = \sqrt{(1 - (-1))^2 + (-2 - 2)^2 + (5 - 1)^2} = \sqrt{2^2 + (-4)^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$
$BC = \sqrt{(4 - 1)^2 + (-7 - (-2))^2 + (8 - 5)^2} = \sqrt{3^2 + (-5)^2 + 3^2} = \sqrt{9 + 25 + 9} = \sqrt{43}$
$CD = \sqrt{(2 - 4)^2 + (-3 - (-7))^2 + (4 - 8)^2} = \sqrt{(-2)^2 + 4^2 + (-4)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$
$DA = \sqrt{(-1 - 2)^2 + (2 - (-3))^2 + (1 - 4)^2} = \sqrt{(-3)^2 + 5^2 + (-3)^2} = \sqrt{9 + 25 + 9} = \sqrt{43}$
Since $AB = CD = 6$ and $BC = DA = \sqrt{43}$,the opposite sides are equal.
Additionally,we check the diagonals $AC$ and $BD$:
$AC = \sqrt{(4 - (-1))^2 + (-7 - 2)^2 + (8 - 1)^2} = \sqrt{5^2 + (-9)^2 + 7^2} = \sqrt{25 + 81 + 49} = \sqrt{155}$
$BD = \sqrt{(2 - 1)^2 + (-3 - (-2))^2 + (4 - 5)^2} = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$
Since the opposite sides are equal and the diagonals are not equal,the quadrilateral $ABCD$ is a parallelogram.

To show that $ABCD$ is a parallelogram,we need to show that opposite sides are equal.
$AB = \sqrt{(-1-1)^{2} + (-2-2)^{2} + (-1-3)^{2}} = \sqrt{(-2)^{2} + (-4)^{2} + (-4)^{2}} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$
$BC = \sqrt{(2 - (-1))^{2} + (3 - (-2))^{2} + (2 - (-1))^{2}} = \sqrt{3^{2} + 5^{2} + 3^{2}} = \sqrt{9 + 25 + 9} = \sqrt{43}$
$CD = \sqrt{(4-2)^{2} + (7-3)^{2} + (6-2)^{2}} = \sqrt{2^{2} + 4^{2} + 4^{2}} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$
$DA = \sqrt{(1-4)^{2} + (2-7)^{2} + (3-6)^{2}} = \sqrt{(-3)^{2} + (-5)^{2} + (-3)^{2}} = \sqrt{9 + 25 + 9} = \sqrt{43}$
Since $AB = CD$ and $BC = DA$,$ABCD$ is a parallelogram.
Now,to prove that $ABCD$ is not a rectangle,we show that the diagonals $AC$ and $BD$ are unequal.
$AC = \sqrt{(2-1)^{2} + (3-2)^{2} + (2-3)^{2}} = \sqrt{1^{2} + 1^{2} + (-1)^{2}} = \sqrt{1 + 1 + 1} = \sqrt{3}$
$BD = \sqrt{(4 - (-1))^{2} + (7 - (-2))^{2} + (6 - (-1))^{2}} = \sqrt{5^{2} + 9^{2} + 7^{2}} = \sqrt{25 + 81 + 49} = \sqrt{155}$
Since $AC \neq BD$,$ABCD$ is not a rectangle.

(A) Let the coordinates of $C$ be $(x, y, z)$.
Given that the centroid $G$ is $(1,1,1)$,and the vertices are $A(3,-5,7)$ and $B(-1,7,-6)$.
The formula for the centroid of a triangle is $G = (\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3})$.
Substituting the values,we get:
$\frac{x+3-1}{3} = 1 \implies x+2 = 3 \implies x = 1$.
$\frac{y-5+7}{3} = 1 \implies y+2 = 3 \implies y = 1$.
$\frac{z+7-6}{3} = 1 \implies z+1 = 3 \implies z = 2$.
Thus,the coordinates of point $C$ are $(1,1,2)$.

(A) Let the coordinates of the fourth vertex $D$ be $(x, y, z)$.
In a parallelogram $ABCD$,the diagonals $AC$ and $BD$ bisect each other at the same midpoint.
Midpoint of $AC = \left(\frac{3 + (-1)}{2}, \frac{-1 + 1}{2}, \frac{2 + 2}{2}\right) = \left(\frac{2}{2}, \frac{0}{2}, \frac{4}{2}\right) = (1, 0, 2)$.
Midpoint of $BD = \left(\frac{x + 1}{2}, \frac{y + 2}{2}, \frac{z - 4}{2}\right)$.
Equating the midpoints:
$\frac{x + 1}{2} = 1$ $\Rightarrow x + 1 = 2$ $\Rightarrow x = 1$.
$\frac{y + 2}{2} = 0$ $\Rightarrow y + 2 = 0$ $\Rightarrow y = -2$.
$\frac{z - 4}{2} = 2$ $\Rightarrow z - 4 = 4$ $\Rightarrow z = 8$.
Thus,the coordinates of the fourth vertex $D$ are $(1, -2, 8)$.

(A) Let $AD, BE,$ and $CF$ be the medians of the given triangle $ABC$ with vertices $A(0,0,6)$,$B(0,4,0)$,and $C(6,0,0)$.
Since $AD$ is the median,$D$ is the mid-point of $BC$.
Coordinates of point $D = \left(\frac{0+6}{2}, \frac{4+0}{2}, \frac{0+0}{2}\right) = (3,2,0)$.
$AD = \sqrt{(0-3)^{2} + (0-2)^{2} + (6-0)^{2}} = \sqrt{9+4+36} = \sqrt{49} = 7$.
Since $BE$ is the median,$E$ is the mid-point of $AC$.
Coordinates of point $E = \left(\frac{0+6}{2}, \frac{0+0}{2}, \frac{6+0}{2}\right) = (3,0,3)$.
$BE = \sqrt{(3-0)^{2} + (0-4)^{2} + (3-0)^{2}} = \sqrt{9+16+9} = \sqrt{34}$.
Since $CF$ is the median,$F$ is the mid-point of $AB$.
Coordinates of point $F = \left(\frac{0+0}{2}, \frac{0+4}{2}, \frac{6+0}{2}\right) = (0,2,3)$.
$CF = \sqrt{(6-0)^{2} + (0-2)^{2} + (0-3)^{2}} = \sqrt{36+4+9} = \sqrt{49} = 7$.
Thus,the lengths of the medians of $\Delta ABC$ are $7, \sqrt{34},$ and $7$.

(A) The coordinates of the centroid of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ 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 the vertices $P(2a, 2, 6)$,$Q(-4, 3b, -10)$,and $R(8, 14, 2c)$,the centroid is:
$\left(\frac{2a - 4 + 8}{3}, \frac{2 + 3b + 14}{3}, \frac{6 - 10 + 2c}{3}\right) = \left(\frac{2a + 4}{3}, \frac{3b + 16}{3}, \frac{2c - 4}{3}\right)$.
Since the origin $(0, 0, 0)$ is the centroid,we equate the coordinates:
$\frac{2a + 4}{3} = 0$ $\Rightarrow 2a = -4$ $\Rightarrow a = -2$.
$\frac{3b + 16}{3} = 0$ $\Rightarrow 3b = -16$ $\Rightarrow b = -\frac{16}{3}$.
$\frac{2c - 4}{3} = 0$ $\Rightarrow 2c = 4$ $\Rightarrow c = 2$.
Thus,the values are $a = -2, b = -\frac{16}{3}, c = 2$.

(N/A) Let the coordinates of the fourth vertex $D$ be $(x, y, z)$.
In a parallelogram,the diagonals bisect each other,meaning they share the same midpoint $P$.
Midpoint of diagonal $AC$ is $P\left(\frac{6-2}{2}, \frac{-2+2}{2}, \frac{4+4}{2}\right) = P(2, 0, 4)$.
Midpoint of diagonal $BD$ is $P\left(\frac{x+2}{2}, \frac{y+4}{2}, \frac{z-8}{2}\right)$.
Equating the midpoints:
$\frac{x+2}{2} = 2$ $\Rightarrow x+2 = 4$ $\Rightarrow x = 2$
$\frac{y+4}{2} = 0$ $\Rightarrow y+4 = 0$ $\Rightarrow y = -4$
$\frac{z-8}{2} = 4$ $\Rightarrow z-8 = 8$ $\Rightarrow z = 16$
Thus,the coordinates of the fourth vertex $D$ are $(2, -4, 16)$.

(A) Let the third vertex of the triangle be $A(x, y, z)$.
The given vertices are $B(2, 4, 6)$ and $C(0, -2, -5)$.
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 $G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
Given that the centroid is the origin $(0, 0, 0)$,we have:
$(0, 0, 0) = \left(\frac{2+0+x}{3}, \frac{4-2+y}{3}, \frac{6-5+z}{3}\right)$.
Equating the coordinates:
$\frac{2+x}{3} = 0 \implies x = -2$.
$\frac{2+y}{3} = 0 \implies y = -2$.
$\frac{1+z}{3} = 0 \implies z = -1$.
Thus,the third vertex is $(-2, -2, -1)$.

(A) 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 the sides are given as $D(1, 2, -3)$,$E(3, 0, 1)$,and $F(-1, 1, -4)$.
We know that the centroid of the triangle formed by the mid-points of the sides of a triangle is the same as the centroid of the original triangle.
The centroid $G(x, y, z)$ of $\Delta DEF$ is calculated as:
$x = \frac{1 + 3 + (-1)}{3} = \frac{3}{3} = 1$
$y = \frac{2 + 0 + 1}{3} = \frac{3}{3} = 1$
$z = \frac{-3 + 1 + (-4)}{3} = \frac{-6}{3} = -2$
Therefore,the centroid of the triangle is $G(1, 1, -2)$.

(N/A) 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 mid-points are $D(5,7,11)$ on $BC$,$E(0,8,5)$ on $AC$,and $F(2,3,-1)$ on $AB$.
Using the mid-point formula:
$1) \frac{x_2+x_3}{2} = 5, \frac{y_2+y_3}{2} = 7, \frac{z_2+z_3}{2} = 11 \Rightarrow x_2+x_3=10, y_2+y_3=14, z_2+z_3=22$
$2) \frac{x_1+x_3}{2} = 0, \frac{y_1+y_3}{2} = 8, \frac{z_1+z_3}{2} = 5 \Rightarrow x_1+x_3=0, y_1+y_3=16, z_1+z_3=10$
$3) \frac{x_1+x_2}{2} = 2, \frac{y_1+y_2}{2} = 3, \frac{z_1+z_2}{2} = -1 \Rightarrow x_1+x_2=4, y_1+y_2=6, z_1+z_2=-2$
Adding all equations for $x$:
$(x_2+x_3) + (x_1+x_3) + (x_1+x_2) = 10+0+4 = 14$ $\Rightarrow 2(x_1+x_2+x_3) = 14$ $\Rightarrow x_1+x_2+x_3 = 7$
$x_1 = (x_1+x_2+x_3) - (x_2+x_3) = 7-10 = -3$
$x_2 = (x_1+x_2+x_3) - (x_1+x_3) = 7-0 = 7$
$x_3 = (x_1+x_2+x_3) - (x_1+x_2) = 7-4 = 3$
Similarly for $y$:
$y_1+y_2+y_3 = (14+16+6)/2 = 18$
$y_1 = 18-14 = 4, y_2 = 18-16 = 2, y_3 = 18-6 = 12$
Similarly for $z$:
$z_1+z_2+z_3 = (22+10-2)/2 = 15$
$z_1 = 15-22 = -7, z_2 = 15-10 = 5, z_3 = 15-(-2) = 17$
Thus,the vertices are $A(-3, 4, -7)$,$B(7, 2, 5)$,and $C(3, 12, 17)$.

(N/A) In a parallelogram,the diagonals bisect each other. Therefore,the mid-point of diagonal $AC$ is the same as the mid-point of diagonal $BD$.
Mid-point of $AC = \left(\frac{1+2}{2}, \frac{2+3}{2}, \frac{3+2}{2}\right) = \left(\frac{3}{2}, \frac{5}{2}, \frac{5}{2}\right)$
Mid-point of $BD = \left(\frac{x-1}{2}, \frac{y-2}{2}, \frac{z-1}{2}\right)$
Equating the mid-points:
$\frac{x-1}{2} = \frac{3}{2}$ $\Rightarrow x-1 = 3$ $\Rightarrow x = 4$
$\frac{y-2}{2} = \frac{5}{2}$ $\Rightarrow y-2 = 5$ $\Rightarrow y = 7$
$\frac{z-1}{2} = \frac{5}{2}$ $\Rightarrow z-1 = 5$ $\Rightarrow z = 6$
Thus,the coordinates of the fourth vertex $D$ are $(4, 7, 6)$.

(A) The vertices of triangle $ABC$ are $A(a, 1, 3)$,$B(-2, b, -5)$,and $C(4, 7, c)$.
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 $G(0, 0, 0)$,we have:
$G(0, 0, 0) = \left(\frac{a-2+4}{3}, \frac{1+b+7}{3}, \frac{3-5+c}{3}\right)$.
Equating the coordinates:
$0 = \frac{a+2}{3} \implies a+2 = 0 \implies a = -2$.
$0 = \frac{b+8}{3} \implies b+8 = 0 \implies b = -8$.
$0 = \frac{c-2}{3} \implies c-2 = 0 \implies c = 2$.
Thus,the values are $a = -2, b = -8, c = 2$.

(N/A) Let the coordinates of $D$ be $(x, y, z)$.
First,calculate the lengths of sides $AB$ and $AC$:
$AB = \sqrt{(5-2)^2 + (6-2)^2 + (9 - (-3))^2} = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$
$AC = \sqrt{(2-2)^2 + (7-2)^2 + (9 - (-3))^2} = \sqrt{0^2 + 5^2 + 12^2} = \sqrt{0 + 25 + 144} = \sqrt{169} = 13$
Since $AB = AC = 13$,$\triangle ABC$ is an isosceles triangle.
In an isosceles triangle,the angle bisector of the vertex angle is also the median to the base.
Therefore,$AD$ is the median to $BC$,which means $D$ is the midpoint of $BC$.
The coordinates of $D$ are given by the midpoint formula:
$D = \left(\frac{5+2}{2}, \frac{6+7}{2}, \frac{9+9}{2}\right) = \left(\frac{7}{2}, \frac{13}{2}, 9\right)$

(N/A) Let the vertices of the triangle be $A(x_1, y_1, z_1), B(x_2, y_2, z_2)$ and $C(x_3, y_3, z_3)$.
Let the mid-points be $D(1, 5, -1)$ on $BC$,$E(0, 4, -2)$ on $AC$,and $F(2, 3, 4)$ on $AB$.
Using the mid-point formula:
$\frac{x_1+x_2}{2} = 2, \frac{y_1+y_2}{2} = 3, \frac{z_1+z_2}{2} = 4 \implies x_1+x_2 = 4, y_1+y_2 = 6, z_1+z_2 = 8$ $(i)$
$\frac{x_1+x_3}{2} = 0, \frac{y_1+y_3}{2} = 4, \frac{z_1+z_3}{2} = -2 \implies x_1+x_3 = 0, y_1+y_3 = 8, z_1+z_3 = -4$ (ii)
$\frac{x_2+x_3}{2} = 1, \frac{y_2+y_3}{2} = 5, \frac{z_2+z_3}{2} = -1 \implies x_2+x_3 = 2, y_2+y_3 = 10, z_2+z_3 = -2$ (iii)
Adding $(i)$,(ii),and (iii): $2(x_1+x_2+x_3) = 6 \implies x_1+x_2+x_3 = 3$. Similarly,$y_1+y_2+y_3 = 12$ and $z_1+z_2+z_3 = 1$.
Subtracting (iii) from the sum: $x_1 = 3-2 = 1, y_1 = 12-10 = 2, z_1 = 1-(-2) = 3$. So,$A = (1, 2, 3)$.
Subtracting (ii) from the sum: $x_2 = 3-0 = 3, y_2 = 12-8 = 4, z_2 = 1-(-4) = 5$. So,$B = (3, 4, 5)$.
Subtracting $(i)$ from the sum: $x_3 = 3-4 = -1, y_3 = 12-6 = 6, z_3 = 1-8 = -7$. So,$C = (-1, 6, -7)$.
Centroid $G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right) = \left(\frac{3}{3}, \frac{12}{3}, \frac{1}{3}\right) = (1, 4, 1/3)$.

(A) Let the given points be $A(-1, -1, 2), B(2, m, 5)$ and $C(3, 11, 6)$.
Since the points are collinear,the vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$ must be parallel.
$\overrightarrow{AB} = (2 - (-1))\hat{i} + (m - (-1))\hat{j} + (5 - 2)\hat{k} = 3\hat{i} + (m + 1)\hat{j} + 3\hat{k}$.
$\overrightarrow{BC} = (3 - 2)\hat{i} + (11 - m)\hat{j} + (6 - 5)\hat{k} = 1\hat{i} + (11 - m)\hat{j} + 1\hat{k}$.
Since $\overrightarrow{AB} = k\overrightarrow{BC}$,we have:
$3 = k(1) \Rightarrow k = 3$.
Equating the $\hat{j}$ components: $m + 1 = k(11 - m)$.
Substituting $k = 3$: $m + 1 = 3(11 - m)$.
$m + 1 = 33 - 3m$.
$4m = 32$.
$m = 8$.

(C) Let the side length of the square base be $x$ and the height of the cuboid be $y$,where $x, y \in \mathbb{Z}^+$.
The sum of all edges of the cuboid is $4x + 4x + 4y = 8x + 4y$.
The sum of the areas of all six faces is $2x^2 + 4xy$.
According to the problem,the sum of edges equals the sum of areas:
$8x + 4y = 2x^2 + 4xy$
Dividing by $2$:
$4x + 2y = x^2 + 2xy$
Rearranging to solve for $y$:
$2y - 2xy = x^2 - 4x$
$2y(1 - x) = x(x - 4)$
$y = \frac{x(x - 4)}{2(1 - x)} = \frac{x(4 - x)}{2(x - 1)}$
Since $y > 0$,we must have $1 < x < 4$. Thus,$x$ can be $2$ or $3$.
If $x = 2$,$y = \frac{2(4 - 2)}{2(2 - 1)} = \frac{4}{2} = 2$.
Sum of edges $= 8(2) + 4(2) = 16 + 8 = 24$.
If $x = 3$,$y = \frac{3(4 - 3)}{2(3 - 1)} = \frac{3}{4}$,which is not an integer.
Thus,the only integer solution is $x = 2, y = 2$,and the sum of edges is $24$.

(B) Since $A(x, y, z)$ lies in the $xy$-plane,$z = 0$. Thus,$A = (x, y, 0)$.
Given $AP^2 = AQ^2 = AR^2$.
$AR^2 = x^2 + y^2 + (0 - 1)^2 = x^2 + y^2 + 1$.
$AP^2 = x^2 + (y - 3)^2 + (0 - 2)^2 = x^2 + y^2 - 6y + 9 + 4 = x^2 + y^2 - 6y + 13$.
$AQ^2 = (x - 2)^2 + y^2 + (0 - 3)^2 = x^2 - 4x + 4 + y^2 + 9 = x^2 + y^2 - 4x + 13$.
Equating $AP^2 = AR^2 \implies x^2 + y^2 - 6y + 13 = x^2 + y^2 + 1 \implies 6y = 12 \implies y = 2$.
Equating $AQ^2 = AR^2 \implies x^2 + y^2 - 4x + 13 = x^2 + y^2 + 1 \implies 4x = 12 \implies x = 3$.
So,$A = (3, 2, 0)$.
Now,calculate the side lengths of $\triangle ABC$ with $A(3, 2, 0)$,$B(1, 4, -1)$,and $C(2, 0, -2)$:
$AB^2 = (3 - 1)^2 + (2 - 4)^2 + (0 - (-1))^2 = 2^2 + (-2)^2 + 1^2 = 4 + 4 + 1 = 9 \implies AB = 3$.
$AC^2 = (3 - 2)^2 + (2 - 0)^2 + (0 - (-2))^2 = 1^2 + 2^2 + 2^2 = 1 + 4 + 4 = 9 \implies AC = 3$.
$BC^2 = (1 - 2)^2 + (4 - 0)^2 + (-1 - (-2))^2 = (-1)^2 + 4^2 + 1^2 = 1 + 16 + 1 = 18 \implies BC = \sqrt{18} = 3\sqrt{2}$.
Since $AB = AC = 3$ and $AB^2 + AC^2 = 9 + 9 = 18 = BC^2$,$\triangle ABC$ is an isosceles right-angled triangle. Thus,$(S1)$ is true.
Area of $\triangle ABC = \frac{1}{2} \times AB \times AC = \frac{1}{2} \times 3 \times 3 = \frac{9}{2}$.
Since $\frac{9}{2} \neq \frac{9\sqrt{2}}{2}$,$(S2)$ is false.
Therefore,only $(S1)$ is true.

(D) The area of $\triangle ABC$ is given by the determinant formula: $\text{Area} = \frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|$.
$\text{Area} = \frac{1}{2} |4(1 - (-3)) + 1(-3 - (-2)) + 9(-2 - 1)|$.
$\text{Area} = \frac{1}{2} |4(4) + 1(-1) + 9(-3)| = \frac{1}{2} |16 - 1 - 27| = \frac{1}{2} |-12| = 6$ square units.
For a parallelogram $AFDE$ inscribed in a triangle $ABC$ such that $D$ lies on $BC$,$E$ on $AC$,and $F$ on $AB$,the maximum area of the parallelogram is exactly half the area of the triangle $ABC$.
$\text{Maximum Area} = \frac{1}{2} \times \text{Area}(\triangle ABC) = \frac{1}{2} \times 6 = 3$ square units.

(B) 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 the formula:
$G = (\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3})$
Given vertices are $P(1, -2, 1)$,$Q(2, 3, -1)$,and $R(1, -1, -1)$.
Substituting the values:
$x$-coordinate $= \frac{1 + 2 + 1}{3} = \frac{4}{3}$
$y$-coordinate $= \frac{-2 + 3 - 1}{3} = \frac{0}{3} = 0$
$z$-coordinate $= \frac{1 - 1 - 1}{3} = -\frac{1}{3}$
Thus,the centroid is $(\frac{4}{3}, 0, -\frac{1}{3})$.

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

(A) To determine if the points $A(1, 1, 2), B(2, 1, 2)$,and $C(2, 2, 1)$ are collinear,we calculate the vectors $\vec{AB}$ and $\vec{BC}$.
$\vec{AB} = (2-1)\hat{i} + (1-1)\hat{j} + (2-2)\hat{k} = 1\hat{i} + 0\hat{j} + 0\hat{k} = \hat{i}$.
$\vec{BC} = (2-2)\hat{i} + (2-1)\hat{j} + (1-2)\hat{k} = 0\hat{i} + 1\hat{j} - 1\hat{k} = \hat{j} - \hat{k}$.
Since $\vec{AB}$ and $\vec{BC}$ are not parallel (they are not scalar multiples of each other),the points are not collinear.
Three non-collinear points in space always form the vertices of a triangle.
Therefore,$A, B, C$ are vertices of a triangle.

(A) Given vertices are $A(-4, 5, p)$,$B(3, 1, 4)$,and $C(-2, 0, q)$.
The centroid $G(r, q, 1)$ 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 $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
Equating the coordinates,we have:
$r = \frac{-4+3-2}{3} = \frac{-3}{3} = -1$
$q = \frac{5+1+0}{3} = \frac{6}{3} = 2$
$1 = \frac{p+4+q}{3} \Rightarrow p+4+q = 3$
Substituting $q=2$ into the equation: $p+4+2 = 3$ $\Rightarrow p+6 = 3$ $\Rightarrow p = -3$.
Now,calculating $2p + q - r$:
$2(-3) + 2 - (-1) = -6 + 2 + 1 = -3$.

(B) The centroid $G$ of a triangle with vertices $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$ is given by $G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
Given $A(x, 4, -1)$,$B(3, x, -5)$,$C(2, -2, 3)$,and $G(2, 1, -1)$.
Equating the $x$-coordinates: $\frac{x+3+2}{3} = 2$.
$x + 5 = 6$.
$x = 1$.
Checking with $y$-coordinates: $\frac{4+x-2}{3} = 1$ $\Rightarrow 2+x = 3$ $\Rightarrow x = 1$.
Thus,the value of $x$ is $1$.

(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 $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.
Given $G = (r, -\frac{4}{3}, \frac{1}{3})$,we equate the coordinates:
$r = \frac{5+1+1}{3} = \frac{7}{3}$
$-\frac{4}{3} = \frac{1+q-2}{3}$ $\Rightarrow -4 = q-1$ $\Rightarrow q = -3$
$\frac{1}{3} = \frac{p+p+3}{3}$ $\Rightarrow 1 = 2p+3$ $\Rightarrow 2p = -2$ $\Rightarrow p = -1$
Thus,the values are $p = -1, q = -3, r = \frac{7}{3}$.

(D) The centroid $G(x, y, z)$ of a triangle with vertices $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$ 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 $G(4, 3, 3)$,$A(a, 3, 1)$,$B(4, 5, b)$,and $C(6, c, 5)$,we have:
$\frac{a+4+6}{3} = 4$ $\Rightarrow a+10 = 12$ $\Rightarrow a = 2$.
$\frac{3+5+c}{3} = 3$ $\Rightarrow 8+c = 9$ $\Rightarrow c = 1$.
$\frac{1+b+5}{3} = 3$ $\Rightarrow 6+b = 9$ $\Rightarrow b = 3$.
Thus,$a=2, b=3, c=1$.

(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 $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3})$.
Given $A(7, -8, 1)$,$B(p, q, 5)$,$C(q+1, 5p, 0)$,and $G(3, -5, r)$.
Equating the coordinates:
$3 = \frac{7 + p + q + 1}{3} \implies 9 = 8 + p + q \implies p + q = 1 \quad (1)$
$-5 = \frac{-8 + q + 5p}{3} \implies -15 = -8 + q + 5p \implies 5p + q = -7 \quad (2)$
$r = \frac{1 + 5 + 0}{3} = \frac{6}{3} = 2$
Subtracting $(1)$ from $(2)$:
$(5p + q) - (p + q) = -7 - 1$
$4p = -8 \implies p = -2$
Substituting $p = -2$ in $(1)$:
$-2 + q = 1 \implies q = 3$
Thus,$p = -2, q = 3, r = 2$.

(C) The vertices are $A(0,4,0)$,$B(0,0,3)$,and $C(0,4,3)$.
Calculate the side lengths:
$a = BC = \sqrt{(0-0)^2 + (4-0)^2 + (3-3)^2} = \sqrt{0+16+0} = 4$.
$b = AC = \sqrt{(0-0)^2 + (4-4)^2 + (3-0)^2} = \sqrt{0+0+9} = 3$.
$c = AB = \sqrt{(0-0)^2 + (0-4)^2 + (3-0)^2} = \sqrt{0+16+9} = 5$.
The incentre $I(x, y, z)$ is given by the formula:
$I = \left( \frac{ax_A + bx_B + cx_C}{a+b+c}, \frac{ay_A + by_B + cy_C}{a+b+c}, \frac{az_A + bz_B + cz_C}{a+b+c} \right)$.
Substituting the values:
$x = \frac{4(0) + 3(0) + 5(0)}{4+3+5} = 0$.
$y = \frac{4(4) + 3(0) + 5(4)}{4+3+5} = \frac{16+0+20}{12} = \frac{36}{12} = 3$.
$z = \frac{4(0) + 3(3) + 5(3)}{4+3+5} = \frac{0+9+15}{12} = \frac{24}{12} = 2$.
Thus,the incentre is $(0,3,2)$.

(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 $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3})$.
Given that the origin $(0, 0, 0)$ is the centroid:
For the $x$-coordinate: $\frac{2+q-5}{3} = 0$ $\Rightarrow q-3 = 0$ $\Rightarrow q = 3$.
For the $y$-coordinate: $\frac{p-2+1}{3} = 0$ $\Rightarrow p-1 = 0$ $\Rightarrow p = 1$.
For the $z$-coordinate: $\frac{-3+5+r}{3} = 0$ $\Rightarrow r+2 = 0$ $\Rightarrow r = -2$.
Thus,$p=1, q=3, r=-2$.

(B) 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:
$G = \left( \frac{x_1+x_2+x_3+x_4}{4}, \frac{y_1+y_2+y_3+y_4}{4}, \frac{z_1+z_2+z_3+z_4}{4} \right)$
Given the vertices $A(3, -5, x)$,$B(5, 4, 2)$,$C(7, -7, y)$,$D(1, 0, z)$ and centroid $G(4, -2, 2)$,we equate the coordinates:
For the $z$-coordinate:
$\frac{x + 2 + y + z}{4} = 2$
$x + y + z + 2 = 8$
$x + y + z = 8 - 2$
$x + y + z = 6$

(B) Let the coordinates of the third vertex $C$ be $(x, y, z)$.
Since $G$ is the centroid of $\triangle ABC$,the coordinates of $G$ are given by the formula:
$G = \left( \frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3}, \frac{z_A + z_B + z_C}{3} \right)$
Given $A(3, 1, 4)$,$B(-4, 5, -3)$,and $G(-1, 2, 1)$:
$-1 = \frac{3 - 4 + x}{3} \implies -3 = -1 + x \implies x = -2$
$2 = \frac{1 + 5 + y}{3} \implies 6 = 6 + y \implies y = 0$
$1 = \frac{4 - 3 + z}{3} \implies 3 = 1 + z \implies z = 2$
Therefore,the third vertex $C$ is $(-2, 0, 2)$.