Section Formula Questions in English

(C) Using the section formula for internal division,the coordinates of the point dividing the segment joining $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m:n$ are given by $\left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right)$.
Given points are $(a, b)$ and $(5, 7)$ with ratio $2:1$ and the dividing point is $(4, 6)$.
For the $x$-coordinate: $\frac{2(5) + 1(a)}{2 + 1} = 4$ $\Rightarrow \frac{10 + a}{3} = 4$ $\Rightarrow 10 + a = 12$ $\Rightarrow a = 2$.
For the $y$-coordinate: $\frac{2(7) + 1(b)}{2 + 1} = 6$ $\Rightarrow \frac{14 + b}{3} = 6$ $\Rightarrow 14 + b = 18$ $\Rightarrow b = 4$.
Thus,$a = 2$ and $b = 4$.

(A) Let the ratio in which the point divides the line segment be $k : 1$.
Using the section formula for the $x$-coordinate:
$\frac{k(-7) + 1(3)}{k + 1} = \frac{1}{2}$
$-14k + 6 = k + 1$
$5 = 15k$
$k = \frac{5}{15} = \frac{1}{3}$
Since $k$ is positive,the division is internal in the ratio $1 : 3$.

(D) The given points are $(a, b)$,$(c, d)$,and $P = \left( \frac{kc + la}{k + l}, \frac{kd + lb}{k + l} \right)$.
By the section formula,a point dividing the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $m:n$ is given by $\left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right)$.
Here,the point $P$ divides the line segment joining $(a, b)$ and $(c, d)$ in the ratio $k:l$.
Since the point $P$ lies on the line segment connecting $(a, b)$ and $(c, d)$,the three points are collinear.

(A) Given $AB = 3AM$. Since $A-M-B$,we have $AB = AM + MB$.
Substituting $AB$,we get $AM + MB = 3AM$,which implies $MB = 2AM$.
Thus,the ratio $AM : MB = 1 : 2$.
Using the section formula for a point $M(x, y)$ dividing $AB$ in the ratio $m:n = 1:2$,where $A(x_1, y_1) = (2, 4)$ and $B(x_2, y_2) = (4, 2)$:
$x = \frac{mx_2 + nx_1}{m+n} = \frac{1(4) + 2(2)}{1+2} = \frac{4+4}{3} = \frac{8}{3}$.
$y = \frac{my_2 + ny_1}{m+n} = \frac{1(2) + 2(4)}{1+2} = \frac{2+8}{3} = \frac{10}{3}$.
Therefore,the coordinates of $M$ are $\left( \frac{8}{3}, \frac{10}{3} \right)$.

(B) The section formula for a point dividing the line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in the ratio $m:n$ is given by $\left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n} \right)$.
Here,$(x_1, y_1, z_1) = (1, 2, 3)$,$(x_2, y_2, z_2) = (3, -5, 6)$,$m = 3$,and $n = -5$.
The sum of the ratio is $m + n = 3 - 5 = -2$.
Calculating the $x$-coordinate: $x = \frac{3(3) + (-5)(1)}{-2} = \frac{9 - 5}{-2} = \frac{4}{-2} = -2$.
Calculating the $y$-coordinate: $y = \frac{3(-5) + (-5)(2)}{-2} = \frac{-15 - 10}{-2} = \frac{-25}{-2} = \frac{25}{2}$.
Calculating the $z$-coordinate: $z = \frac{3(6) + (-5)(3)}{-2} = \frac{18 - 15}{-2} = \frac{3}{-2} = -\frac{3}{2}$.
Thus,the point is $\left( -2, \frac{25}{2}, -\frac{3}{2} \right)$,which matches option $B$.

(A) The coordinates of a point $P$ dividing the line segment joining $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ externally in the ratio $m:n$ are given by the formula:
$P = \left( \frac{mx_2 - nx_1}{m - n}, \frac{my_2 - ny_1}{m - n}, \frac{mz_2 - nz_1}{m - n} \right)$
Given $A(1, 2, -1)$,$B(-1, 0, 1)$,$m = 1$,and $n = 2$.
Substituting the values:
$x = \frac{1(-1) - 2(1)}{1 - 2} = \frac{-1 - 2}{-1} = \frac{-3}{-1} = 3$
$y = \frac{1(0) - 2(2)}{1 - 2} = \frac{0 - 4}{-1} = \frac{-4}{-1} = 4$
$z = \frac{1(1) - 2(-1)}{1 - 2} = \frac{1 + 2}{-1} = \frac{3}{-1} = -3$
Thus,the coordinates of $P$ are $(3, 4, -3)$.

(A) Let the point $C(5, 4, -6)$ divide the line segment joining $A(9, 8, -10)$ and $B(3, 2, -4)$ in the ratio $k:1$.
Using the section formula,the coordinates of $C$ are given by:
$C = \left( \frac{3k + 9}{k + 1}, \frac{2k + 8}{k + 1}, \frac{-4k - 10}{k + 1} \right)$
Equating the $x$-coordinate to $5$:
$\frac{3k + 9}{k + 1} = 5$
$3k + 9 = 5k + 5$
$4 = 2k$
$k = 2$
Thus,the ratio $k:1$ is $2:1$.
Therefore,the point $C$ divides the line segment $AB$ in the ratio $2:1$.

(D) Let the points be $A(2, -1, 3)$ and $B(4, 3, 1)$. The ratio is $m_1 : m_2 = 3 : 4$.
Using the section formula for internal division:
$x = \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2} = \frac{3(4) + 4(2)}{3 + 4} = \frac{12 + 8}{7} = \frac{20}{7}$
$y = \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} = \frac{3(3) + 4(-1)}{3 + 4} = \frac{9 - 4}{7} = \frac{5}{7}$
$z = \frac{m_1 z_2 + m_2 z_1}{m_1 + m_2} = \frac{3(1) + 4(3)}{3 + 4} = \frac{3 + 12}{7} = \frac{15}{7}$
Thus,the coordinates are $(\frac{20}{7}, \frac{5}{7}, \frac{15}{7})$.
Therefore,the correct option is $D$.

(D) Let the ratio in which the $xy$-plane divides the line segment joining the points $P(a, b, c)$ and $Q(-a, -c, -b)$ be $k:1$.
The coordinates of the point dividing the segment in ratio $k:1$ are given by the section formula:
$\left( \frac{k(-a) + 1(a)}{k+1}, \frac{k(-c) + 1(b)}{k+1}, \frac{k(-b) + 1(c)}{k+1} \right)$
Since the point lies on the $xy$-plane,its $z$-coordinate must be $0$:
$\frac{-kb + c}{k+1} = 0$
This implies $-kb + c = 0$,or $kb = c$.
Therefore,$k = \frac{c}{b}$.
Thus,the required ratio is $c:b$.

(C) Let the ratio in which the $yz$-plane divides the line segment joining the points $A(2, 4, 5)$ and $B(3, 5, -4)$ be $k:1$.
By the section formula,the coordinates of the point dividing the segment are given by $\left( \frac{3k+2}{k+1}, \frac{5k+4}{k+1}, \frac{-4k+5}{k+1} \right)$.
Since the point lies on the $yz$-plane,its $x$-coordinate must be zero.
Therefore,$\frac{3k+2}{k+1} = 0$.
This implies $3k + 2 = 0$,so $k = -\frac{2}{3}$.
Thus,the ratio is $-\frac{2}{3}:1$,which is $-2:3$.

(B) Let $AD$ be the angle bisector of $\angle A$,where $D$ lies on $BC$. The point $D$ divides $BC$ in the ratio $AB : AC$.
First,calculate the lengths of sides $AB$ and $AC$:
$\vec{AB} = (2-4, 3-7, 4-8) = (-2, -4, -4) \implies |\vec{AB}| = \sqrt{(-2)^2 + (-4)^2 + (-4)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$.
$\vec{AC} = (2-4, 5-7, 7-8) = (-2, -2, -1) \implies |\vec{AC}| = \sqrt{(-2)^2 + (-2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
The ratio $AB : AC = 6 : 3 = 2 : 1$.
Using the section formula,the coordinates of $D$ are:
$D = \frac{1 \cdot B + 2 \cdot C}{1 + 2} = \frac{(2, 3, 4) + 2(2, 5, 7)}{3} = \frac{(2+4, 3+10, 4+14)}{3} = \frac{(6, 13, 18)}{3} = (2, \frac{13}{3}, 6)$.
Now,the length of the angle bisector $AD$ is the distance between $A(4, 7, 8)$ and $D(2, \frac{13}{3}, 6)$:
$|AD| = \sqrt{(2-4)^2 + (\frac{13}{3} - 7)^2 + (6-8)^2} = \sqrt{(-2)^2 + (\frac{13-21}{3})^2 + (-2)^2} = \sqrt{4 + (-\frac{8}{3})^2 + 4} = \sqrt{8 + \frac{64}{9}} = \sqrt{\frac{72+64}{9}} = \sqrt{\frac{136}{9}} = \frac{\sqrt{4 \times 34}}{3} = \frac{2\sqrt{34}}{3}$.

(B) Let the points be $A(2, 4, 5)$ and $B(3, 5, -4)$. The ratio is $m : n = -2 : 3$.
Using the section formula,the coordinates of the point $P(x, y, z)$ are given by:
$x = \frac{mx_2 + nx_1}{m + n} = \frac{(-2)(3) + (3)(2)}{-2 + 3} = \frac{-6 + 6}{1} = 0$
$y = \frac{my_2 + ny_1}{m + n} = \frac{(-2)(5) + (3)(4)}{-2 + 3} = \frac{-10 + 12}{1} = 2$
$z = \frac{mz_2 + nz_1}{m + n} = \frac{(-2)(-4) + (3)(5)}{-2 + 3} = \frac{8 + 15}{1} = 23$
The coordinates of the point are $(0, 2, 23)$.
Since the $x$-coordinate is $0$,the point lies on the $YOZ$ plane.

(A) The formula for the coordinates of a point dividing the line segment joining $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ externally in the ratio $m : n$ is given by:
$\left( \frac{mx_2 - nx_1}{m - n}, \frac{my_2 - ny_1}{m - n}, \frac{mz_2 - nz_1}{m - n} \right)$
Here,$A = (1, 2, -1)$,$B = (-1, 0, 1)$,$m = 1$,and $n = 2$.
Substituting the values:
$x = \frac{1(-1) - 2(1)}{1 - 2} = \frac{-1 - 2}{-1} = \frac{-3}{-1} = 3$
$y = \frac{1(0) - 2(2)}{1 - 2} = \frac{0 - 4}{-1} = \frac{-4}{-1} = 4$
$z = \frac{1(1) - 2(-1)}{1 - 2} = \frac{1 + 2}{-1} = \frac{3}{-1} = -3$
Thus,the coordinates are $(3, 4, -3)$.

(A) Let the ratio in which the $yz$-plane divides the line segment joining the points $A(-2, 4, 7)$ and $B(3, -5, 8)$ be $k : 1$.
The coordinates of the point dividing the segment in ratio $k : 1$ are given by the section formula:
$P = \left( \frac{k(3) + 1(-2)}{k+1}, \frac{k(-5) + 1(4)}{k+1}, \frac{k(8) + 1(7)}{k+1} \right)$
Since the point $P$ lies on the $yz$-plane,its $x$-coordinate must be $0$:
$\frac{3k - 2}{k + 1} = 0$
Solving for $k$:
$3k - 2 = 0$
$3k = 2$
$k = \frac{2}{3}$
Thus,the ratio is $k : 1 = \frac{2}{3} : 1$,which is $2 : 3$.

(C) The coordinates of point $A$ are given by the section formula: $A = \left( \frac{3k - 5}{k + 1}, \frac{5k + 1}{k + 1} \right)$.
The area of $\Delta ABC$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 2$.
Substituting the coordinates $A\left( \frac{3k - 5}{k + 1}, \frac{5k + 1}{k + 1} \right)$,$B(1, 5)$,and $C(7, -2)$:
$\frac{1}{2} \left| \frac{3k - 5}{k + 1}(5 - (-2)) + 1(-2 - \frac{5k + 1}{k + 1}) + 7(\frac{5k + 1}{k + 1} - 5) \right| = 2$.
Simplifying the expression inside the modulus:
$\frac{1}{2} \left| \frac{7(3k - 5) - (2(k + 1) + 5k + 1) + 7(5k + 1 - 5k - 5)}{k + 1} \right| = 2$.
$\frac{1}{2} \left| \frac{21k - 35 - 7k - 3 - 28}{k + 1} \right| = 2$.
$\frac{1}{2} \left| \frac{14k - 66}{k + 1} \right| = 2$.
$|14k - 66| = 4|k + 1|$.
Case $1$: $14k - 66 = 4k + 4$ $\Rightarrow 10k = 70$ $\Rightarrow k = 7$.
Case $2$: $14k - 66 = -4k - 4$ $\Rightarrow 18k = 62$ $\Rightarrow k = 31/9$.
Thus,$k = 7$ or $k = 31/9$.

(C) Let the coordinates of point $B$ be $(x, y)$.
Given that point $P(5, -1)$ divides the segment $AB$ in the ratio $m:n = 2:3$.
Using the section formula,the coordinates of $P$ are given by:
$P = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right)$
Substituting the values:
$5 = \frac{2(x) + 3(11)}{2+3} \implies 5 = \frac{2x + 33}{5} \implies 25 = 2x + 33 \implies 2x = -8 \implies x = -4$
$-1 = \frac{2(y) + 3(-3)}{2+3} \implies -1 = \frac{2y - 9}{5} \implies -5 = 2y - 9 \implies 2y = 4 \implies y = 2$
Therefore,the coordinates of point $B$ are $(-4, 2)$.

(A) Let the $x$-axis divide the line segment joining the points $A(3, 4)$ and $B(5, 6)$ in the ratio $k:1$ at point $P(x, 0)$.
Using the section formula,the $y$-coordinate of point $P$ is given by:
$y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2}$
Since the point lies on the $x$-axis,its $y$-coordinate is $0$.
$0 = \frac{k(6) + 1(4)}{k + 1}$
$0 = 6k + 4$
$6k = -4$
$k = -\frac{4}{6} = -\frac{2}{3}$
Thus,the ratio is $-2:3$ or $-\frac{2}{3}$.

(A) Let the ratio be $k : 1$. Using the section formula,the coordinates of the point dividing the segment joining $(x_1, y_1)$ and $(x_2, y_2)$ in the ratio $k : 1$ are given by $(\frac{kx_2 + x_1}{k+1}, \frac{ky_2 + y_1}{k+1})$.
Given points are $(8, 4)$ and $(9, 6)$,and the dividing point is $(5, -2)$.
Equating the $x$-coordinate: $\frac{9k + 8}{k+1} = 5$.
$9k + 8 = 5k + 5$.
$4k = -3$.
$k = -\frac{3}{4}$.
Since $k$ is negative,the division is external in the ratio $3 : 4$.

(A) Let $P(x, y, z)$ be the point which divides the line segment joining $A(1, -2, 3)$ and $B(3, 4, -5)$ internally in the ratio $m:n = 2:3$.
Using the section formula:
$x = \frac{mx_2 + nx_1}{m+n} = \frac{2(3) + 3(1)}{2+3} = \frac{6+3}{5} = \frac{9}{5}$
$y = \frac{my_2 + ny_1}{m+n} = \frac{2(4) + 3(-2)}{2+3} = \frac{8-6}{5} = \frac{2}{5}$
$z = \frac{mz_2 + nz_1}{m+n} = \frac{2(-5) + 3(3)}{2+3} = \frac{-10+9}{5} = -\frac{1}{5}$
Thus,the required point is $\left(\frac{9}{5}, \frac{2}{5}, -\frac{1}{5}\right)$.

(A) Let the points be $A(1, -2, 3)$ and $B(3, 4, -5)$. The point $P(x, y, z)$ divides the line segment $AB$ externally in the ratio $m:n = 2:3$.
The section formula for external division is given by:
$x = \frac{mx_2 - nx_1}{m - n}, y = \frac{my_2 - ny_1}{m - n}, z = \frac{mz_2 - nz_1}{m - n}$
Substituting the values $m=2, n=3, x_1=1, y_1=-2, z_1=3, x_2=3, y_2=4, z_2=-5$:
$x = \frac{2(3) - 3(1)}{2 - 3} = \frac{6 - 3}{-1} = -3$
$y = \frac{2(4) - 3(-2)}{2 - 3} = \frac{8 + 6}{-1} = -14$
$z = \frac{2(-5) - 3(3)}{2 - 3} = \frac{-10 - 9}{-1} = 19$
Thus,the coordinates of the point are $(-3, -14, 19)$.

(N/A) Let the points be $A(-4, 6, 10)$,$B(2, 4, 6)$,and $C(14, 0, -2)$.
Assume that a point $P$ divides the line segment $AB$ in the ratio $k:1$.
Using the section formula,the coordinates of $P$ are given by:
$P = \left( \frac{2k - 4}{k + 1}, \frac{4k + 6}{k + 1}, \frac{6k + 10}{k + 1} \right)$
For the points to be collinear,point $C$ must coincide with point $P$ for some value of $k$.
Equating the $x$-coordinates: $\frac{2k - 4}{k + 1} = 14$
$2k - 4 = 14k + 14 \implies -12k = 18 \implies k = -\frac{18}{12} = -\frac{3}{2}$
Now,check the $y$-coordinate for $k = -\frac{3}{2}$:
$\frac{4(-\frac{3}{2}) + 6}{-\frac{3}{2} + 1} = \frac{-6 + 6}{-\frac{1}{2}} = 0$
Now,check the $z$-coordinate for $k = -\frac{3}{2}$:
$\frac{6(-\frac{3}{2}) + 10}{-\frac{3}{2} + 1} = \frac{-9 + 10}{-\frac{1}{2}} = \frac{1}{-\frac{1}{2}} = -2$
Since the coordinates match point $C(14, 0, -2)$,the point $C$ lies on the line passing through $A$ and $B$.
Thus,the points $A$,$B$,and $C$ are collinear.

(B) Let $A, B, C$ be the vertices of the triangle with coordinates $(x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2})$ and $(x_{3}, y_{3}, z_{3})$ respectively.
Let $D$ be the midpoint of $BC$. The coordinates of $D$ are $\left(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}, \frac{z_{2}+z_{3}}{2}\right)$.
The centroid $G$ divides the median $AD$ in the ratio $2:1$.
Using the section formula,the coordinates of $G$ are $\left(\frac{2\left(\frac{x_{2}+x_{3}}{2}\right) + x_{1}}{2+1}, \frac{2\left(\frac{y_{2}+y_{3}}{2}\right) + y_{1}}{2+1}, \frac{2\left(\frac{z_{2}+z_{3}}{2}\right) + z_{1}}{2+1}\right)$.
Simplifying this,we get $\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}, \frac{z_{1}+z_{2}+z_{3}}{3}\right)$.

(B) Let the $YZ$-plane divide the line segment joining $A(4, 8, 10)$ and $B(6, 10, -8)$ at point $P$ in the ratio $k:1$.
Using the section formula,the coordinates of $P$ are given by $\left(\frac{6k+4}{k+1}, \frac{10k+8}{k+1}, \frac{-8k+10}{k+1}\right)$.
Since $P$ lies on the $YZ$-plane,its $x$-coordinate must be $0$.
Therefore,$\frac{6k+4}{k+1} = 0$.
This implies $6k + 4 = 0$,so $k = -\frac{4}{6} = -\frac{2}{3}$.
The negative sign indicates that the division is external.
Thus,the $YZ$-plane divides the line segment in the ratio $2:3$ externally.

(A) The coordinates of a point $R$ that divides the line segment joining points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ internally in the ratio $m:n$ are given by the section formula:
$\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n}\right)$
Given points are $P(-2, 3, 5)$ and $Q(1, -4, 6)$ with ratio $m:n = 2:3$.
Substituting the values:
$x = \frac{2(1) + 3(-2)}{2+3} = \frac{2 - 6}{5} = -\frac{4}{5}$
$y = \frac{2(-4) + 3(3)}{2+3} = \frac{-8 + 9}{5} = \frac{1}{5}$
$z = \frac{2(6) + 3(5)}{2+3} = \frac{12 + 15}{5} = \frac{27}{5}$
Thus,the coordinates of the required point are $\left(-\frac{4}{5}, \frac{1}{5}, \frac{27}{5}\right)$.

(A) The coordinates of point $R$ that divides the line segment joining points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ externally in the ratio $m:n$ are given by the formula:
$\left(\frac{mx_2 - nx_1}{m-n}, \frac{my_2 - ny_1}{m-n}, \frac{mz_2 - nz_1}{m-n}\right)$
Given points are $P(-2, 3, 5)$ and $Q(1, -4, 6)$ with ratio $m:n = 2:3$.
Substituting the values:
$x = \frac{2(1) - 3(-2)}{2-3} = \frac{2 + 6}{-1} = -8$
$y = \frac{2(-4) - 3(3)}{2-3} = \frac{-8 - 9}{-1} = \frac{-17}{-1} = 17$
$z = \frac{2(6) - 3(5)}{2-3} = \frac{12 - 15}{-1} = \frac{-3}{-1} = 3$
Thus,the coordinates of the required point are $(-8, 17, 3)$.

(A) Let point $Q(5, 4, -6)$ divide the line segment joining points $P(3, 2, -4)$ and $R(9, 8, -10)$ in the ratio $k: 1$.
By the section formula,the coordinates of $Q$ are given by:
$Q = \left( \frac{k(9) + 3}{k + 1}, \frac{k(8) + 2}{k + 1}, \frac{k(-10) - 4}{k + 1} \right)$
Equating the $x$-coordinates:
$\frac{9k + 3}{k + 1} = 5$
$9k + 3 = 5k + 5$
$4k = 2$
$k = \frac{2}{4} = \frac{1}{2}$
Thus,the ratio $k: 1$ is $\frac{1}{2}: 1$,which is $1: 2$.
Therefore,point $Q$ divides $PR$ in the ratio $1: 2$.

(A) Let the $YZ$-plane divide the line segment joining the points $A(-2, 4, 7)$ and $B(3, -5, 8)$ in the ratio $k: 1$.
By the section formula,the coordinates of the point of intersection $P$ are given by $\left(\frac{3k - 2}{k + 1}, \frac{-5k + 4}{k + 1}, \frac{8k + 7}{k + 1}\right)$.
Since the point $P$ lies on the $YZ$-plane,its $x$-coordinate must be $0$.
Therefore,$\frac{3k - 2}{k + 1} = 0$.
This implies $3k - 2 = 0$,so $k = \frac{2}{3}$.
Thus,the ratio is $k: 1 = \frac{2}{3}: 1$,which is $2: 3$.

(N/A) Let the point $C$ divide the line segment $AB$ in the ratio $k:1$.
By the section formula,the coordinates of the point dividing $AB$ are $\left(\frac{k(-1) + 2}{k+1}, \frac{k(2) - 3}{k+1}, \frac{k(1) + 4}{k+1}\right)$.
Equating these to the coordinates of point $C(0, \frac{1}{3}, 2)$,we have:
$\frac{-k+2}{k+1} = 0 \implies -k+2 = 0 \implies k = 2$.
Now,check the other coordinates for $k=2$:
$y$-coordinate: $\frac{2(2)-3}{2+1} = \frac{4-3}{3} = \frac{1}{3}$.
$z$-coordinate: $\frac{2(1)+4}{2+1} = \frac{6}{3} = 2$.
Since the coordinates match for $k=2$,point $C$ lies on the line segment $AB$ dividing it in the ratio $2:1$.
Therefore,the points $A, B,$ and $C$ are collinear.

(A) Let $A$ and $B$ be the points that trisect the line segment joining points $P(4, 2, -6)$ and $Q(10, -16, 6)$.
Point $A$ divides $PQ$ in the ratio $1:2$. Therefore,by the section formula,the coordinates of point $A$ are given by:
$A = \left(\frac{1(10) + 2(4)}{1 + 2}, \frac{1(-16) + 2(2)}{1 + 2}, \frac{1(6) + 2(-6)}{1 + 2}\right) = \left(\frac{18}{3}, \frac{-12}{3}, \frac{-6}{3}\right) = (6, -4, -2)$
Point $B$ divides $PQ$ in the ratio $2:1$. Therefore,by the section formula,the coordinates of point $B$ are given by:
$B = \left(\frac{2(10) + 1(4)}{2 + 1}, \frac{2(-16) + 1(2)}{2 + 1}, \frac{2(6) + 1(-6)}{2 + 1}\right) = \left(\frac{24}{3}, \frac{-30}{3}, \frac{6}{3}\right) = (8, -10, 2)$
Thus,the coordinates of the points are $(6, -4, -2)$ and $(8, -10, 2)$.

(A) The coordinates of points $P$ and $Q$ are given as $P(2, -3, 4)$ and $Q(8, 0, 10)$.
Let $R$ divide the line segment $PQ$ in the ratio $k:1$.
By the section formula,the coordinates of point $R$ are given by $\left(\frac{8k+2}{k+1}, \frac{-3}{k+1}, \frac{10k+4}{k+1}\right)$.
It is given that the $x$-coordinate of point $R$ is $4$.
$\therefore \frac{8k+2}{k+1} = 4$
$\Rightarrow 8k + 2 = 4k + 4$
$\Rightarrow 4k = 2$
$\Rightarrow k = \frac{1}{2}$.
Substituting $k = \frac{1}{2}$ into the coordinates of $R$:
$y$-coordinate: $\frac{-3}{\frac{1}{2} + 1} = \frac{-3}{\frac{3}{2}} = -2$.
$z$-coordinate: $\frac{10(\frac{1}{2}) + 4}{\frac{1}{2} + 1} = \frac{5 + 4}{\frac{3}{2}} = \frac{9}{\frac{3}{2}} = 6$.
Thus,the coordinates of point $R$ are $(4, -2, 6)$.

(N/A) Let $P(x_{1}, y_{1}, z_{1})$ and $Q(x_{2}, y_{2}, z_{2})$ be the points that trisect the line segment $AB$.
Since point $P$ divides $AB$ in the ratio $1:2$ internally,we have:
$P = \left(\frac{1(5) + 2(2)}{1 + 2}, \frac{1(-8) + 2(1)}{1 + 2}, \frac{1(3) + 2(-3)}{1 + 2}\right)$
$P = \left(\frac{5 + 4}{3}, \frac{-8 + 2}{3}, \frac{3 - 6}{3}\right)$
$P = \left(\frac{9}{3}, \frac{-6}{3}, \frac{-3}{3}\right) = (3, -2, -1)$
Since point $Q$ divides $AB$ in the ratio $2:1$ internally,we have:
$Q = \left(\frac{2(5) + 1(2)}{2 + 1}, \frac{2(-8) + 1(1)}{2 + 1}, \frac{2(3) + 1(-3)}{2 + 1}\right)$
$Q = \left(\frac{10 + 2}{3}, \frac{-16 + 1}{3}, \frac{6 - 3}{3}\right)$
$Q = \left(\frac{12}{3}, \frac{-15}{3}, \frac{3}{3}\right) = (4, -5, 1)$
Thus,the coordinates of the points are $(3, -2, -1)$ and $(4, -5, 1)$.

(D) Let the point $P$ divide the line segment $QR$ in the ratio $\lambda : 1$.
Using the section formula,the coordinates of $P$ are given by:
$P = \left( \frac{5\lambda + 2}{\lambda + 1}, \frac{\lambda + 2}{\lambda + 1}, \frac{-2\lambda + 1}{\lambda + 1} \right)$.
Given that the $x-$coordinate of $P$ is $4$,we have:
$\frac{5\lambda + 2}{\lambda + 1} = 4$.
Solving for $\lambda$:
$5\lambda + 2 = 4(\lambda + 1) \Rightarrow 5\lambda + 2 = 4\lambda + 4 \Rightarrow \lambda = 2$.
Now,substitute $\lambda = 2$ into the expression for the $z-$coordinate:
$z = \frac{-2(2) + 1}{2 + 1} = \frac{-4 + 1}{3} = \frac{-3}{3} = -1$.
Thus,the $z-$coordinate of the point is $-1$.

(2:3) The given points are $A (1,-2,-8), B (5,0,-2)$ and $C (11,3,7)$.
$\overrightarrow{AB} = (5-1)\hat{i} + (0+2)\hat{j} + (-2+8)\hat{k} = 4\hat{i} + 2\hat{j} + 6\hat{k}$.
$\overrightarrow{BC} = (11-5)\hat{i} + (3-0)\hat{j} + (7+2)\hat{k} = 6\hat{i} + 3\hat{j} + 9\hat{k}$.
$\overrightarrow{AC} = (11-1)\hat{i} + (3+2)\hat{j} + (7+8)\hat{k} = 10\hat{i} + 5\hat{j} + 15\hat{k}$.
$|\overrightarrow{AB}| = \sqrt{4^2 + 2^2 + 6^2} = \sqrt{16+4+36} = \sqrt{56} = 2\sqrt{14}$.
$|\overrightarrow{BC}| = \sqrt{6^2 + 3^2 + 9^2} = \sqrt{36+9+81} = \sqrt{126} = 3\sqrt{14}$.
$|\overrightarrow{AC}| = \sqrt{10^2 + 5^2 + 15^2} = \sqrt{100+25+225} = \sqrt{350} = 5\sqrt{14}$.
Since $|\overrightarrow{AC}| = |\overrightarrow{AB}| + |\overrightarrow{BC}|$,the points $A, B,$ and $C$ are collinear.
Let point $B$ divide $AC$ in the ratio $\lambda : 1$. Using the section formula:
$\vec{b} = \frac{\lambda\vec{c} + 1\vec{a}}{\lambda + 1}$.
$5\hat{i} + 0\hat{j} - 2\hat{k} = \frac{\lambda(11\hat{i} + 3\hat{j} + 7\hat{k}) + (1\hat{i} - 2\hat{j} - 8\hat{k})}{\lambda + 1}$.
Equating the $\hat{j}$ components:
$0 = \frac{3\lambda - 2}{\lambda + 1} \Rightarrow 3\lambda - 2 = 0 \Rightarrow \lambda = \frac{2}{3}$.
Thus,the ratio is $2:3$.

(D) Let the point $C$ divide the line segment $AB$ in the ratio $k: 1$.
Using the section formula,the coordinates of $C$ are given by:
$C = \left( \frac{k(2) + 1(1)}{k+1}, \frac{k(-1) + 1(0)}{k+1}, \frac{k(3) + 1(1)}{k+1} \right)$
Given $C = (3, -2, 5)$,we equate the $x$-coordinates:
$\frac{2k + 1}{k+1} = 3$
$2k + 1 = 3k + 3$
$2k - 3k = 3 - 1$
$-k = 2$
$k = -2$
Thus,the ratio is $-2: 1$.

(D) The points of trisection divide the line segment $AB$ in the ratio $1:2$ and $2:1$ internally.
Let the points be $P$ and $Q$ such that $P$ divides $AB$ in ratio $1:2$ and $Q$ divides $AB$ in ratio $2:1$.
The $z$-coordinate of a point dividing the segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in ratio $m:n$ is given by $\frac{mz_2 + nz_1}{m+n}$.
For $z_1$ (ratio $1:2$):
$z_1 = \frac{1(6) + 2(4)}{1 + 2} = \frac{6 + 8}{3} = \frac{14}{3}$.
For $z_2$ (ratio $2:1$):
$z_2 = \frac{2(6) + 1(4)}{2 + 1} = \frac{12 + 4}{3} = \frac{16}{3}$.
Therefore,$z_1 + z_2 = \frac{14}{3} + \frac{16}{3} = \frac{30}{3} = 10$.

(A) Since $AD$ is the angle bisector of $\angle A$,by the Angle Bisector Theorem,we have $\frac{AB}{AC} = \frac{BD}{CD}$.
First,calculate the lengths of sides $AB$ and $AC$:
$AB = \sqrt{(5-7)^2 + (4-6)^2 + (6-4)^2} = \sqrt{(-2)^2 + (-2)^2 + 2^2} = \sqrt{4+4+4} = \sqrt{12} = 2\sqrt{3}$.
$AC = \sqrt{(3-7)^2 + (2-6)^2 + (0-4)^2} = \sqrt{(-4)^2 + (-4)^2 + (-4)^2} = \sqrt{16+16+16} = \sqrt{48} = 4\sqrt{3}$.
Thus,the ratio $\frac{BD}{CD} = \frac{AB}{AC} = \frac{2\sqrt{3}}{4\sqrt{3}} = \frac{1}{2}$.
Point $D$ divides the line segment $BC$ internally in the ratio $1:2$. Using the section formula,the coordinates of $D$ are:
$D = \left(\frac{1(3) + 2(5)}{1+2}, \frac{1(2) + 2(4)}{1+2}, \frac{1(0) + 2(6)}{1+2}\right)$
$D = \left(\frac{3+10}{3}, \frac{2+8}{3}, \frac{0+12}{3}\right) = \left(\frac{13}{3}, \frac{10}{3}, 4\right)$.

(A) Let the points be $A(3,-2,2)$ and $B(6,-17,-4)$. Let $P(2,3,4)$ divide the line segment $AB$ in the ratio $\lambda : 1$.
Using the section formula for the $x$-coordinate:
$2 = \frac{6\lambda + 3}{\lambda + 1}$
$2\lambda + 2 = 6\lambda + 3$
$-4\lambda = 1 \implies \lambda = -\frac{1}{4}$.
The harmonic conjugate $Q$ of $P$ with respect to $A$ and $B$ divides $AB$ externally in the same ratio $\lambda : 1$,which is equivalent to dividing $AB$ internally in the ratio $-\lambda : 1$.
Thus,$Q$ divides $AB$ in the ratio $\frac{1}{4} : 1$,or $1 : 4$.
Using the section formula for $Q$ with ratio $m:n = 1:4$:
$Q = \left( \frac{1(6) + 4(3)}{1+4}, \frac{1(-17) + 4(-2)}{1+4}, \frac{1(-4) + 4(2)}{1+4} \right)$
$Q = \left( \frac{6+12}{5}, \frac{-17-8}{5}, \frac{-4+8}{5} \right)$
$Q = \left( \frac{18}{5}, -\frac{25}{5}, \frac{4}{5} \right) = \left( \frac{18}{5}, -5, \frac{4}{5} \right)$.

(D) Let $O$ be the orthocenter $(3, -4, 2)$ and $C$ be the circumcenter $(2, 1, 3)$.
We know that the centroid $G$ divides the line segment joining the orthocenter and the circumcenter in the ratio $2:1$.
Using the section formula,the coordinates of the centroid $G$ are given by:
$G = \left(\frac{1(3) + 2(2)}{1+2}, \frac{1(-4) + 2(1)}{1+2}, \frac{1(2) + 2(3)}{1+2}\right)$
$G = \left(\frac{3+4}{3}, \frac{-4+2}{3}, \frac{2+6}{3}\right)$
$G = \left(\frac{7}{3}, \frac{-2}{3}, \frac{8}{3}\right)$

(A) Let the point $P$ with abscissa $3$ divide the line segment joining $A(6, 5)$ and $B(-1, 4)$ in the ratio $\lambda: 1$.
Using the section formula for the $x$-coordinate:
$x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2}$
$3 = \frac{\lambda(-1) + 1(6)}{\lambda + 1}$
$3(\lambda + 1) = -\lambda + 6$
$3\lambda + 3 = -\lambda + 6$
$4\lambda = 3$
$\lambda = \frac{3}{4}$
Thus,the ratio $\lambda: 1$ is $3: 4$.
Hence,option $A$ is correct.

(B) Let the points be $A(3,-2,2)$ and $B(6,-17,-4)$. Let $P(2,3,4)$ divide $AB$ in the ratio $k:1$.
Using the section formula: $(2,3,4) = \left(\frac{6k+3}{k+1}, \frac{-17k-2}{k+1}, \frac{-4k+2}{k+1}\right)$.
Equating the $x$-coordinates: $2 = \frac{6k+3}{k+1}$ $\Rightarrow 2k+2 = 6k+3$ $\Rightarrow -4k = 1$ $\Rightarrow k = -\frac{1}{4}$.
The harmonic conjugate $Q$ divides $AB$ in the ratio $-k:1$,which is $\frac{1}{4}:1$ or $1:4$ externally.
The coordinates of $Q$ are given by $\left(\frac{1(6)+4(3)}{1+4}, \frac{1(-17)+4(-2)}{1+4}, \frac{1(-4)+4(2)}{1+4}\right)$.
Calculating these: $Q = \left(\frac{6+12}{5}, \frac{-17-8}{5}, \frac{-4+8}{5}\right) = \left(\frac{18}{5}, -5, \frac{4}{5}\right)$.

(B) Let the required point be $P$. By the section formula,the coordinates of point $P$ dividing the line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in the ratio $m:n$ are given by:
$P = \left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n}\right)$
Here,$(x_1, y_1, z_1) = (3, -2, 1)$,$(x_2, y_2, z_2) = (-2, 3, 11)$,$m = 2$,and $n = 3$.
Substituting these values into the formula:
$P = \left(\frac{2(-2) + 3(3)}{2+3}, \frac{2(3) + 3(-2)}{2+3}, \frac{2(11) + 3(1)}{2+3}\right)$
$P = \left(\frac{-4 + 9}{5}, \frac{6 - 6}{5}, \frac{22 + 3}{5}\right)$
$P = \left(\frac{5}{5}, \frac{0}{5}, \frac{25}{5}\right)$
$P = (1, 0, 5)$

(D) Given $P=(3,12,4)$ and $O=(0,0,0)$.
The distance $OP = \sqrt{3^2 + 12^2 + 4^2} = \sqrt{9 + 144 + 16} = \sqrt{169} = 13$.
Since $Q$ lies on $OP$ and $OQ=3$,the point $Q$ divides the segment $OP$ in the ratio $3 : (13-3) = 3 : 10$ internally or $3 : -10$ externally.
The coordinates of $Q$ are given by the section formula: $Q = \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}, \frac{m z_2 + n z_1}{m+n} \right)$.
For internal division $(3:10)$: $Q = \left( \frac{3 \times 3}{13}, \frac{3 \times 12}{13}, \frac{3 \times 4}{13} \right) = \left( \frac{9}{13}, \frac{36}{13}, \frac{12}{13} \right)$.
Sum of coordinates $= \frac{9+36+12}{13} = \frac{57}{13}$.
For external division $(3:-10)$: $Q = \left( \frac{3 \times 3}{-7}, \frac{3 \times 12}{-7}, \frac{3 \times 4}{-7} \right) = \left( -\frac{9}{7}, -\frac{36}{7}, -\frac{12}{7} \right)$.
However,checking the options,the intended answer is $\frac{57}{13}$.

(B) First,calculate the lengths of the sides of $\triangle ABC$:
$AB = \sqrt{(2-1)^2 + (0-2)^2 + (1-0)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$
$AC = \sqrt{(-3-1)^2 + (0-2)^2 + (2-0)^2} = \sqrt{16 + 4 + 4} = \sqrt{24} = 2\sqrt{6}$
$BC = \sqrt{(-3-2)^2 + (0-0)^2 + (2-1)^2} = \sqrt{25 + 0 + 1} = \sqrt{26}$
Since $AD$ is the internal angle bisector of $\angle BAC$,it divides the opposite side $BC$ in the ratio of the adjacent sides $AB:AC = \sqrt{6}:2\sqrt{6} = 1:2$.
Using the section formula,the coordinates of point $D$ are:
$D = \left( \frac{1(-3) + 2(2)}{1+2}, \frac{1(0) + 2(0)}{1+2}, \frac{1(2) + 2(1)}{1+2} \right) = \left( \frac{1}{3}, 0, \frac{4}{3} \right)$
The length of $AD$ is:
$AD = \sqrt{(1 - 1/3)^2 + (2 - 0)^2 + (0 - 4/3)^2} = \sqrt{(2/3)^2 + 2^2 + (-4/3)^2} = \sqrt{4/9 + 4 + 16/9} = \sqrt{20/9 + 36/9} = \sqrt{56/9} = \frac{\sqrt{56}}{3} = \frac{2\sqrt{14}}{3}$

(A) Let the points be $A(2, -5, 3)$ and $B(-1, -8, 0)$. The point $P(0, -7, 1)$ divides the line segment $AB$ in the ratio $k:1$.
Using the section formula:
$0 = \frac{k(-1) + 1(2)}{k+1} \implies -k + 2 = 0 \implies k = 2$.
So,$P$ divides $AB$ internally in the ratio $2:1$.
Since $Q$ is the harmonic conjugate of $P$ with respect to $AB$,$Q$ divides $AB$ externally in the ratio $2:1$.
Using the external section formula for $Q(\alpha, \beta, \gamma)$:
$\alpha = \frac{2(-1) - 1(2)}{2-1} = -2 - 2 = -4$.
$\beta = \frac{2(-8) - 1(-5)}{2-1} = -16 + 5 = -11$.
$\gamma = \frac{2(0) - 1(3)}{2-1} = 0 - 3 = -3$.
Thus,$Q = (-4, -11, -3)$.
We need to find $\alpha - \beta + \gamma = -4 - (-11) + (-3) = -4 + 11 - 3 = 4$.

(A) Given points are $P(\alpha, 4, 7)$ and $Q(3, \beta, 8)$.
Since the $YZ$-plane divides the line segment $PQ$ in the ratio $2:3$,the $x$-coordinate of the point of division is zero.
Using the section formula: $\frac{2(3) + 3(\alpha)}{2+3} = 0 \Rightarrow 6 + 3\alpha = 0 \Rightarrow \alpha = -2$.
Since the $ZX$-plane divides the line segment $PQ$ in the ratio $4:5$,the $y$-coordinate of the point of division is zero.
Using the section formula: $\frac{4(\beta) + 5(4)}{4+5} = 0 \Rightarrow 4\beta + 20 = 0 \Rightarrow \beta = -5$.
Thus,the points are $P(-2, 4, 7)$ and $Q(3, -5, 8)$.
The length of the line segment $PQ$ is given by the distance formula:
$PQ = \sqrt{(3 - (-2))^2 + (-5 - 4)^2 + (8 - 7)^2}$
$PQ = \sqrt{(5)^2 + (-9)^2 + (1)^2}$
$PQ = \sqrt{25 + 81 + 1} = \sqrt{107}$.

(A) Let the point $C$ divide the line segment $AB$ externally in the ratio $k: 1$.
The coordinates of point $C$ are given by the section formula for external division:
$C = \left( \frac{k x_2 - x_1}{k - 1}, \frac{k y_2 - y_1}{k - 1}, \frac{k z_2 - z_1}{k - 1} \right)$
Substituting the coordinates $A(1, 2, 3)$ and $B(3, 4, 7)$:
$C = \left( \frac{3k - 1}{k - 1}, \frac{4k - 2}{k - 1}, \frac{7k - 3}{k - 1} \right)$
Given $C = (-3, -2, -5)$,we equate the $x$-coordinates:
$\frac{3k - 1}{k - 1} = -3$
$3k - 1 = -3(k - 1)$
$3k - 1 = -3k + 3$
$6k = 4$
$k = \frac{4}{6} = \frac{2}{3}$
Thus,the ratio $k: 1$ is $\frac{2}{3}: 1$,which is $2: 3$.

(B) The formula for the coordinates of a point dividing the line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in the ratio $m:n$ externally is given by:
$\left(\frac{mx_2 - nx_1}{m - n}, \frac{my_2 - ny_1}{m - n}, \frac{mz_2 - nz_1}{m - n}\right)$
Given points are $(2, 3, 4)$ and $(3, -4, 7)$ with ratio $m:n = 2:4$.
Substituting the values:
$x = \frac{2(3) - 4(2)}{2 - 4} = \frac{6 - 8}{-2} = \frac{-2}{-2} = 1$
$y = \frac{2(-4) - 4(3)}{2 - 4} = \frac{-8 - 12}{-2} = \frac{-20}{-2} = 10$
$z = \frac{2(7) - 4(4)}{2 - 4} = \frac{14 - 16}{-2} = \frac{-2}{-2} = 1$
Thus,the coordinates are $(1, 10, 1)$.

(B) Given that the point $(a, 8, -2)$ divides the line segment joining $(1, 4, 6)$ and $(5, 2, 10)$ in the ratio $m: n$.
By the section formula,the coordinates are given by:
$x = \frac{5m + n}{m + n}$,$y = \frac{2m + 4n}{m + n}$,$z = \frac{10m + 6n}{m + n}$.
Equating the $y$-coordinate:
$8 = \frac{2m + 4n}{m + n} \implies 8m + 8n = 2m + 4n \implies 6m = -4n \implies \frac{m}{n} = -\frac{2}{3}$.
Equating the $z$-coordinate:
$-2 = \frac{10m + 6n}{m + n} \implies -2m - 2n = 10m + 6n \implies -12m = 8n \implies \frac{m}{n} = -\frac{8}{12} = -\frac{2}{3}$.
Since both equations yield the same ratio,we find $a$ using the $x$-coordinate:
$a = \frac{5m + n}{m + n} = \frac{5(\frac{m}{n}) + 1}{(\frac{m}{n}) + 1} = \frac{5(-\frac{2}{3}) + 1}{(-\frac{2}{3}) + 1} = \frac{-\frac{10}{3} + 1}{\frac{1}{3}} = \frac{-\frac{7}{3}}{\frac{1}{3}} = -7$.
Finally,calculating the expression:
$\frac{2m}{n} - \frac{a}{3} = 2(-\frac{2}{3}) - (\frac{-7}{3}) = -\frac{4}{3} + \frac{7}{3} = \frac{3}{3} = 1$.

(B) Given points $A(1, 2, -1)$ and $B(-1, 0, 1)$. Point $P$ divides the line segment $AB$ externally in the ratio $m_1: m_2 = 1: 2$.
For external division,the coordinates of $P$ are given by $\left( \frac{m_1 x_2 - m_2 x_1}{m_1 - m_2}, \frac{m_1 y_2 - m_2 y_1}{m_1 - m_2}, \frac{m_1 z_2 - m_2 z_1}{m_1 - m_2} \right)$.
Substituting the values:
$P = \left( \frac{1(-1) - 2(1)}{1 - 2}, \frac{1(0) - 2(2)}{1 - 2}, \frac{1(1) - 2(-1)}{1 - 2} \right)$
$P = \left( \frac{-1 - 2}{-1}, \frac{-4}{-1}, \frac{1 + 2}{-1} \right)$
$P = \left( \frac{-3}{-1}, \frac{-4}{-1}, \frac{3}{-1} \right) = (3, 4, -3)$.
Now,we find the distance $PQ$ where $Q = (1, 3, -1)$:
$PQ = \sqrt{(3 - 1)^2 + (4 - 3)^2 + (-3 - (-1))^2}$
$PQ = \sqrt{(2)^2 + (1)^2 + (-2)^2}$
$PQ = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$ units.