Section Formula Questions in English

(B) Let the points be $P(1,2,3)$ and $R(4,5,6)$. The point $Q(3,4,5)$ divides $PR$ in the ratio $\lambda: 1$.
Using the section formula,the coordinates of $Q$ are given by:
$Q = \left( \frac{\lambda(4) + 1(1)}{\lambda + 1}, \frac{\lambda(5) + 1(2)}{\lambda + 1}, \frac{\lambda(6) + 1(3)}{\lambda + 1} \right) = (3,4,5)$
Comparing the $x$-coordinates:
$\frac{4\lambda + 1}{\lambda + 1} = 3$
$4\lambda + 1 = 3\lambda + 3$
$\lambda = 2$
Now,we need to find the point that divides the line segment joining $Q(3,4,5)$ and $P(1,2,3)$ in the ratio $-1: \lambda$ (which is $-1: 2$).
Let the required point be $S(x, y, z)$. Using the section formula for internal division with ratio $m:n = -1: 2$:
$x = \frac{-1(1) + 2(3)}{-1 + 2} = \frac{-1 + 6}{1} = 5$
$y = \frac{-1(2) + 2(4)}{-1 + 2} = \frac{-2 + 8}{1} = 6$
$z = \frac{-1(3) + 2(5)}{-1 + 2} = \frac{-3 + 10}{1} = 7$
Thus,the required point is $(5,6,7)$.

(D) Let $A = (3, -2, 2)$ and $B = (6, -17, -4)$. Let $P = (2, 3, 4)$ divide the line segment $AB$ in the ratio $m:n$.
Using the section formula,we have:
$P = \left(\frac{6m + 3n}{m+n}, \frac{-17m - 2n}{m+n}, \frac{-4m + 2n}{m+n}\right) = (2, 3, 4)$.
Equating the $x$-coordinates: $\frac{6m + 3n}{m+n} = 2 \implies 6m + 3n = 2m + 2n \implies 4m = -n \implies \frac{m}{n} = -\frac{1}{4}$.
Thus,$P$ divides $AB$ externally in the ratio $1:4$.
The harmonic conjugate of $P$ with respect to $A$ and $B$ divides the segment $AB$ internally in the ratio $m:n = 1:4$.
Using the section formula for internal division:
$Q = \left(\frac{1(6) + 4(3)}{1+4}, \frac{1(-17) + 4(-2)}{1+4}, \frac{1(-4) + 4(2)}{1+4}\right) = \left(\frac{6+12}{5}, \frac{-17-8}{5}, \frac{-4+8}{5}\right) = \left(\frac{18}{5}, -5, \frac{4}{5}\right)$.
Therefore,the correct option is $D$.

(A) Let the point $R$ divide the line segment $PQ$ in the ratio $\lambda : 1$.
Using the section formula for the $x$-coordinate:
$x = \frac{m x_2 + n x_1}{m + n}$
$-5 = \frac{\lambda(-9) + 1(-3)}{\lambda + 1}$
$-5(\lambda + 1) = -9\lambda - 3$
$-5\lambda - 5 = -9\lambda - 3$
$-5\lambda + 9\lambda = 5 - 3$
$4\lambda = 2$
$\lambda = \frac{2}{4} = \frac{1}{2}$
Thus,the ratio $\lambda : 1$ is $\frac{1}{2} : 1$,which simplifies to $1 : 2$.

(C) Let the point $P(-9, 12, -15)$ divide the line segment joining $A(1, -2, 3)$ and $B(-4, 5, -6)$ in the ratio $\lambda : 1$.
Using the section formula,we have:
$-9 = \frac{-4\lambda + 1}{\lambda + 1} \implies -9\lambda - 9 = -4\lambda + 1 \implies -5\lambda = 10 \implies \lambda = -2$.
The point $P$ divides $AB$ externally in the ratio $2 : 1$.
The harmonic conjugate of $P$ with respect to $AB$ is the point $Q$ that divides $AB$ internally in the same ratio $2 : 1$.
Using the internal section formula:
$Q = \left(\frac{2(-4) + 1(1)}{2+1}, \frac{2(5) + 1(-2)}{2+1}, \frac{2(-6) + 1(3)}{2+1}\right)$
$Q = \left(\frac{-8 + 1}{3}, \frac{10 - 2}{3}, \frac{-12 + 3}{3}\right)$
$Q = \left(-\frac{7}{3}, \frac{8}{3}, -3\right)$.

(C) The harmonic conjugate of a point $A$ with respect to points $B$ and $C$ is a point $P$ that divides the line segment $BC$ externally in the ratio of the distances $AB$ and $AC$.
First,we calculate the distances $AB$ and $AC$:
$AB = \sqrt{(6-5)^2 + (2-4)^2 + (-1-2)^2} = \sqrt{1^2 + (-2)^2 + (-3)^2} = \sqrt{1+4+9} = \sqrt{14}$.
$AC = \sqrt{(8-5)^2 + (-2-4)^2 + (-7-2)^2} = \sqrt{3^2 + (-6)^2 + (-9)^2} = \sqrt{9+36+81} = \sqrt{126} = 3\sqrt{14}$.
The ratio $m:n = AB:AC = \sqrt{14} : 3\sqrt{14} = 1:3$.
Since $P$ divides $BC$ externally in the ratio $1:3$,the coordinates of $P$ are given by the section formula for external division:
$P = \left(\frac{m x_2 - n x_1}{m-n}, \frac{m y_2 - n y_1}{m-n}, \frac{m z_2 - n z_1}{m-n}\right)$
Substituting $m=1, n=3, B=(6,2,-1), C=(8,-2,-7)$:
$x = \frac{1(8) - 3(6)}{1-3} = \frac{8-18}{-2} = \frac{-10}{-2} = 5$.
Wait,let us re-evaluate the definition. The harmonic conjugate $P$ of $A$ with respect to $B$ and $C$ means $A$ divides $BC$ internally in ratio $k$ and $P$ divides $BC$ externally in ratio $k$.
$A$ divides $BC$ in ratio $k:1$.
$5 = \frac{k(8) + 1(6)}{k+1} \implies 5k+5 = 8k+6 \implies 3k = -1 \implies k = -1/3$.
Since $k$ is negative,$A$ divides $BC$ externally.
Using the property that $P$ is the harmonic conjugate,$P$ divides $BC$ internally in the same ratio $1:3$.
$P = \left(\frac{1(8) + 3(6)}{1+3}, \frac{1(-2) + 3(2)}{1+3}, \frac{1(-7) + 3(-1)}{1+3}\right) = \left(\frac{26}{4}, \frac{4}{4}, \frac{-10}{4}\right) = (6.5, 1, -2.5) = (\frac{13}{2}, 1, \frac{-5}{2})$.
Thus,the correct option is $C$.

(A) Let the coordinates of point $B$ be $(x, y, z)$.
Given that point $P(-2, 3, 5)$ divides the line segment joining $A(1, 3, 4)$ and $B(x, y, z)$ in the ratio $m:n = 1:3$.
Using the section formula for internal division:
$P = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n} \right)$
Substituting the values:
$(-2, 3, 5) = \left( \frac{1(x) + 3(1)}{1+3}, \frac{1(y) + 3(3)}{1+3}, \frac{1(z) + 3(4)}{1+3} \right)$
$(-2, 3, 5) = \left( \frac{x+3}{4}, \frac{y+9}{4}, \frac{z+12}{4} \right)$
Equating the coordinates:
$1) \frac{x+3}{4} = -2 \Rightarrow x+3 = -8 \Rightarrow x = -11$
$2) \frac{y+9}{4} = 3 \Rightarrow y+9 = 12 \Rightarrow y = 3$
$3) \frac{z+12}{4} = 5 \Rightarrow z+12 = 20 \Rightarrow z = 8$
Thus,the coordinates of $B$ are $(-11, 3, 8)$.

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

(D) The length of the sides of $\triangle ABC$ are calculated using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.
$c = AB = \sqrt{(2-2)^2 + (5-(-1))^2 + (1-1)^2} = \sqrt{0^2 + 6^2 + 0^2} = 6$.
$b = AC = \sqrt{(0-2)^2 + (-2-(-1))^2 + (3-1)^2} = \sqrt{(-2)^2 + (-1)^2 + 2^2} = \sqrt{4+1+4} = 3$.
$a = BC = \sqrt{(0-2)^2 + (-2-5)^2 + (3-1)^2} = \sqrt{(-2)^2 + (-7)^2 + 2^2} = \sqrt{4+49+4} = \sqrt{57}$.
By the Angle Bisector Theorem,point $D$ divides $BC$ in the ratio $c:b$,which is $6:3 = 2:1$.
Using the section formula,the coordinates of $D$ are $\left(\frac{2(0)+1(2)}{2+1}, \frac{2(-2)+1(5)}{2+1}, \frac{2(3)+1(1)}{2+1}\right) = \left(\frac{2}{3}, \frac{1}{3}, \frac{7}{3}\right)$.
Now,$AD = \sqrt{(\frac{2}{3}-2)^2 + (\frac{1}{3}-(-1))^2 + (\frac{7}{3}-1)^2} = \sqrt{(-\frac{4}{3})^2 + (\frac{4}{3})^2 + (\frac{4}{3})^2} = \sqrt{\frac{16}{9} + \frac{16}{9} + \frac{16}{9}} = \sqrt{\frac{48}{9}} = \frac{4\sqrt{3}}{3} = \frac{4}{\sqrt{3}}$.

(B) The angle bisector theorem states that the bisector of $\angle BAC$ divides the opposite side $BC$ in the ratio of the sides adjacent to the angle,i.e.,$BD/DC = AB/AC$.
First,calculate the lengths of sides $AB$ and $AC$:
$AB = \sqrt{(5-3)^2 + (3-2)^2 + (2-0)^2} = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4+1+4} = \sqrt{9} = 3$.
$AC = \sqrt{(-9-3)^2 + (6-2)^2 + (-3-0)^2} = \sqrt{(-12)^2 + 4^2 + (-3)^2} = \sqrt{144 + 16 + 9} = \sqrt{169} = 13$.
Thus,the ratio $BD:DC = AB:AC = 3:13$.
Using the section formula,the coordinates of $D$ are given by:
$D = \left(\frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}, \frac{m z_2 + n z_1}{m+n}\right)$ where $m=3, n=13$,$B(5,3,2)$,and $C(-9,6,-3)$.
$x = \frac{3(-9) + 13(5)}{3+13} = \frac{-27 + 65}{16} = \frac{38}{16}$.
$y = \frac{3(6) + 13(3)}{3+13} = \frac{18 + 39}{16} = \frac{57}{16}$.
$z = \frac{3(-3) + 13(2)}{3+13} = \frac{-9 + 26}{16} = \frac{17}{16}$.
Therefore,the coordinates of $D$ are $\left(\frac{38}{16}, \frac{57}{16}, \frac{17}{16}\right)$.

(B) Let the required point be $P(x, y, z)$. 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:
$P = \left(\frac{2(-2) + 3(3)}{2+3}, \frac{2(3) + 3(-2)}{2+3}, \frac{2(11) + 3(1)}{2+3}\right)$
$P = \left(\frac{-4 + 9}{5}, \frac{6 - 6}{5}, \frac{22 + 3}{5}\right)$
$P = \left(\frac{5}{5}, \frac{0}{5}, \frac{25}{5}\right)$
$P = (1, 0, 5)$

(A) The angle bisector theorem states that the bisector of $\angle BAC$ divides the opposite side $BC$ in the ratio of the sides adjacent to the angle,i.e.,$BD:CD = AB:AC$.
First,calculate the lengths of sides $AB$ and $AC$:
$AB = \sqrt{(0-4)^2 + (-2-3)^2 + (2-5)^2} = \sqrt{(-4)^2 + (-5)^2 + (-3)^2} = \sqrt{16 + 25 + 9} = \sqrt{50} = 5\sqrt{2}$.
$AC = \sqrt{(3-4)^2 + (2-3)^2 + (1-5)^2} = \sqrt{(-1)^2 + (-1)^2 + (-4)^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}$.
Thus,the ratio $BD:CD = AB:AC = 5\sqrt{2} : 3\sqrt{2} = 5:3$.
Point $D$ divides $BC$ internally in the ratio $5:3$. Using the section formula,the coordinates of $D$ are:
$D = \left(\frac{5(3) + 3(0)}{5+3}, \frac{5(2) + 3(-2)}{5+3}, \frac{5(1) + 3(2)}{5+3}\right)$
$D = \left(\frac{15+0}{8}, \frac{10-6}{8}, \frac{5+6}{8}\right) = \left(\frac{15}{8}, \frac{4}{8}, \frac{11}{8}\right)$.

(C) The lengths of the sides $AB$ and $AC$ are calculated as follows:
$AB = \sqrt{(4-3)^2 + (1-2)^2 + (0-(-1))^2} = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$
$AC = \sqrt{(2-3)^2 + (1-2)^2 + (4-(-1))^2} = \sqrt{(-1)^2 + (-1)^2 + 5^2} = \sqrt{1 + 1 + 25} = \sqrt{27} = 3\sqrt{3}$
By the Angle Bisector Theorem,the bisector of $\angle BAC$ divides the opposite side $BC$ in the ratio of the adjacent sides:
$\frac{BD}{DC} = \frac{AB}{AC} = \frac{\sqrt{3}}{3\sqrt{3}} = \frac{1}{3}$
Using the section formula,the coordinates of $D(p, q, r)$ are:
$D = \left( \frac{1(2) + 3(4)}{1+3}, \frac{1(1) + 3(1)}{1+3}, \frac{1(4) + 3(0)}{1+3} \right) = \left( \frac{14}{4}, \frac{4}{4}, \frac{4}{4} \right) = \left( \frac{7}{2}, 1, 1 \right)$
Thus,$p = \frac{7}{2}, q = 1, r = 1$.
Finally,$\sqrt{2p + q + r} = \sqrt{2(\frac{7}{2}) + 1 + 1} = \sqrt{7 + 1 + 1} = \sqrt{9} = 3$.

(C) Given vertices $A(4,7,8)$,$B(2,3,4)$,and $C(2,5,7)$.
First,calculate the lengths of sides $AB$ and $AC$:
$AB = \sqrt{(4-2)^2 + (7-3)^2 + (8-4)^2} = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{4+16+16} = \sqrt{36} = 6$.
$AC = \sqrt{(4-2)^2 + (7-5)^2 + (8-7)^2} = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4+4+1} = \sqrt{9} = 3$.
By the Angle Bisector Theorem,the internal bisector of angle $A$ divides the opposite side $BC$ in the ratio $AB:AC = 6:3 = 2:1$.
Let $D$ be the point on $BC$ dividing it in the ratio $2:1$. Using the section formula:
$D = \left( \frac{2(2) + 1(2)}{2+1}, \frac{2(5) + 1(3)}{2+1}, \frac{2(7) + 1(4)}{2+1} \right) = \left( \frac{6}{3}, \frac{13}{3}, \frac{18}{3} \right) = \left( 2, \frac{13}{3}, 6 \right)$.
The length of the internal bisector is the distance $AD$:
$AD = \sqrt{(4-2)^2 + (7 - \frac{13}{3})^2 + (8-6)^2} = \sqrt{2^2 + (\frac{8}{3})^2 + 2^2} = \sqrt{4 + \frac{64}{9} + 4} = \sqrt{8 + \frac{64}{9}} = \sqrt{\frac{72+64}{9}} = \sqrt{\frac{136}{9}} = \frac{\sqrt{4 \times 34}}{3} = \frac{2}{3} \sqrt{34}$.

(C) The vertices of the tetrahedron are $O(0,0,0), A(3,1,4), B(1,3,2)$,and $C(0,4,-2)$.
The centroid $G$ of the tetrahedron is given by the average of its vertices: $G = \left(\frac{0+3+1+0}{4}, \frac{0+1+3+4}{4}, \frac{0+4+2-2}{4}\right) = \left(1, 2, 1\right)$.
The centroid $G_1$ of the face $ABC$ is given by the average of vertices $A, B$,and $C$: $G_1 = \left(\frac{3+1+0}{3}, \frac{1+3+4}{3}, \frac{4+2-2}{3}\right) = \left(\frac{4}{3}, \frac{8}{3}, \frac{4}{3}\right)$.
We need to find the point $P$ that divides the line segment $GG_1$ in the ratio $1:2$. Using the section formula,$P = \left(\frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}, \frac{m z_2 + n z_1}{m+n}\right)$ with $m=1, n=2$,$G(1,2,1)$ and $G_1(\frac{4}{3}, \frac{8}{3}, \frac{4}{3})$:
$x = \frac{1(\frac{4}{3}) + 2(1)}{1+2} = \frac{\frac{4}{3} + 2}{3} = \frac{10}{9}$.
$y = \frac{1(\frac{8}{3}) + 2(2)}{1+2} = \frac{\frac{8}{3} + 4}{3} = \frac{20}{9}$.
$z = \frac{1(\frac{4}{3}) + 2(1)}{1+2} = \frac{\frac{4}{3} + 2}{3} = \frac{10}{9}$.
Thus,the point is $\left(\frac{10}{9}, \frac{20}{9}, \frac{10}{9}\right)$.

(A) Let the points be $A(2, p, 2)$ and $B(p, -2, p)$. The point $P\left(\frac{8}{5}, -\frac{1}{5}, \frac{8}{5}\right)$ divides $AB$ in the ratio $m:n$.
Using the section formula,the $y$-coordinate of $P$ is given by:
$\frac{m(-2) + n(p)}{m+n} = -\frac{1}{5}$
$-10m + 5np = -m - n$
$9m = n(5p + 1) \implies \frac{m}{n} = \frac{5p+1}{9}$
Using the $x$-coordinate of $P$:
$\frac{m(p) + n(2)}{m+n} = \frac{8}{5}$
$5mp + 10n = 8m + 8n$
$m(5p - 8) = -2n \implies \frac{m}{n} = \frac{-2}{5p-8} = \frac{2}{8-5p}$
Equating the two ratios:
$\frac{5p+1}{9} = \frac{2}{8-5p}$
$(5p+1)(8-5p) = 18$
$40p - 25p^2 + 8 - 5p = 18$
$25p^2 - 35p + 10 = 0$
$5p^2 - 7p + 2 = 0$
$(5p-2)(p-1) = 0$
Since $p$ is an integer,$p = 1$.
Then $\frac{m}{n} = \frac{5(1)+1}{9} = \frac{6}{9} = \frac{2}{3}$.
We need to find $\frac{3m+n}{3n} = \frac{3(m/n) + 1}{3} = \frac{3(2/3) + 1}{3} = \frac{2+1}{3} = 1$.
Since $p=1$,the result is $p$.

(B) First,calculate the lengths of sides $AB$ and $AC$ using the distance formula:
$AB = \sqrt{(4-1)^2 + (3-(-2))^2 + (5-1)^2} = \sqrt{3^2 + 5^2 + 4^2} = \sqrt{9 + 25 + 16} = \sqrt{50} = 5\sqrt{2}$
$AC = \sqrt{(4-3)^2 + (3-2)^2 + (5-1)^2} = \sqrt{1^2 + 1^2 + 4^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}$
According to the Angle Bisector Theorem,the internal bisector of $\angle BAC$ divides the opposite side $BC$ in the ratio of the sides containing the angle:
$\frac{BD}{DC} = \frac{AB}{AC} = \frac{5\sqrt{2}}{3\sqrt{2}} = \frac{5}{3}$
Thus,point $D$ divides $BC$ internally in the ratio $5:3$. Using the section formula,the coordinates of $D$ are:
$D = \left( \frac{5(3) + 3(1)}{5+3}, \frac{5(2) + 3(-2)}{5+3}, \frac{5(1) + 3(1)}{5+3} \right) = \left( \frac{18}{8}, \frac{4}{8}, \frac{8}{8} \right) = \left( \frac{9}{4}, \frac{1}{2}, 1 \right)$
Now,calculate the length $CD$ using the distance formula between $C(3,2,1)$ and $D(\frac{9}{4}, \frac{1}{2}, 1)$:
$CD = \sqrt{(\frac{9}{4} - 3)^2 + (\frac{1}{2} - 2)^2 + (1 - 1)^2}$
$CD = \sqrt{(-\frac{3}{4})^2 + (-\frac{3}{2})^2 + 0^2} = \sqrt{\frac{9}{16} + \frac{9}{4}} = \sqrt{\frac{9 + 36}{16}} = \sqrt{\frac{45}{16}} = \frac{3\sqrt{5}}{4}$

(C) The coordinates of point $P$ dividing $AB$ in the ratio $m:n = 1:2$ are given by the section formula:
$P = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n} \right)$
$P = \left( \frac{1(2) + 2(1)}{1+2}, \frac{1(3) + 2(2)}{1+2}, \frac{1(1) + 2(3)}{1+2} \right) = \left( \frac{4}{3}, \frac{7}{3}, \frac{7}{3} \right)$
The coordinates of point $Q$ dividing $BC$ in the ratio $m:n = -2:3$ are:
$Q = \left( \frac{-2(3) + 3(2)}{-2+3}, \frac{-2(1) + 3(3)}{-2+3}, \frac{-2(2) + 3(1)}{-2+3} \right) = \left( \frac{0}{1}, \frac{7}{1}, \frac{-1}{1} \right) = (0, 7, -1)$
The distance $PQ$ is given by $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$:
$PQ = \sqrt{\left(0 - \frac{4}{3}\right)^2 + \left(7 - \frac{7}{3}\right)^2 + \left(-1 - \frac{7}{3}\right)^2}$
$PQ = \sqrt{\left(-\frac{4}{3}\right)^2 + \left(\frac{14}{3}\right)^2 + \left(-\frac{10}{3}\right)^2}$
$PQ = \sqrt{\frac{16}{9} + \frac{196}{9} + \frac{100}{9}} = \sqrt{\frac{312}{9}} = \frac{\sqrt{312}}{3} = \frac{2\sqrt{78}}{3}$

(C) Let the point $B$ divide the line segment $AC$ in the ratio $k: 1$. Using the section formula,the coordinates of $B$ are given by:
$B = \left( \frac{k(9) + 1(3)}{k+1}, \frac{k(8) + 1(2)}{k+1}, \frac{k(10) + 1(4)}{k+1} \right)$
Given that $B = \left( \frac{33}{5}, \frac{28}{5}, \frac{38}{5} \right)$,we equate the $x$-coordinates:
$\frac{9k + 3}{k + 1} = \frac{33}{5}$
$5(9k + 3) = 33(k + 1)$
$45k + 15 = 33k + 33$
$45k - 33k = 33 - 15$
$12k = 18$
$k = \frac{18}{12} = \frac{3}{2}$
Thus,the ratio $k: 1$ is $3: 2$.

(B) Given that the points $A(-1,0,7), B(3,2, t)$ and $C(5, k,-2)$ are collinear,the direction ratios of $AB$ and $BC$ must be proportional.
$\frac{3-(-1)}{5-3} = \frac{2-0}{k-2} = \frac{t-7}{-2-t}$
$\frac{4}{2} = \frac{2}{k-2} = \frac{t-7}{-2-t}$
$2 = \frac{2}{k-2} \Rightarrow k-2 = 1 \Rightarrow k = 3$
$2 = \frac{t-7}{-2-t} \Rightarrow -4-2t = t-7 \Rightarrow 3t = 3 \Rightarrow t = 1$
Thus,the points are $B(3,2,1)$ and $C(5,3,-2)$.
The point $P$ is given by $P(t, k-2t, t+k) = P(1, 3-2(1), 1+3) = P(1, 1, 4)$.
Let $P$ divide the line segment $BC$ in the ratio $\lambda : 1$.
Using the section formula for the $x$-coordinate:
$1 = \frac{\lambda(5) + 1(3)}{\lambda + 1}$
$\lambda + 1 = 5\lambda + 3$
$-2 = 4\lambda$
$\lambda = -\frac{1}{2}$
Therefore,the required ratio is $-1: 2$.

(A) The vertices of the triangle are $A(4,3,2), B(5,4,6)$,and $C(-1,-1,5)$.
First,we calculate the lengths of sides $AB$ and $AC$ using the distance formula:
$AB = \sqrt{(5-4)^2 + (4-3)^2 + (6-2)^2} = \sqrt{1^2 + 1^2 + 4^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}$.
$AC = \sqrt{(-1-4)^2 + (-1-3)^2 + (5-2)^2} = \sqrt{(-5)^2 + (-4)^2 + 3^2} = \sqrt{25 + 16 + 9} = \sqrt{50} = 5\sqrt{2}$.
According to the Angle Bisector Theorem,the bisector of angle $A$ divides the opposite side $BC$ in the ratio of the sides containing the angle,which is $AB : AC = 3\sqrt{2} : 5\sqrt{2} = 3 : 5$.
Let $D$ be the point on $BC$ where the bisector meets it. $D$ divides $BC$ internally in the ratio $m:n = 3:5$.
Using the section formula $D = \left(\frac{mx_C + nx_B}{m+n}, \frac{my_C + ny_B}{m+n}, \frac{mz_C + nz_B}{m+n}\right)$:
$x_D = \frac{3(-1) + 5(5)}{3+5} = \frac{-3 + 25}{8} = \frac{22}{8}$.
$y_D = \frac{3(-1) + 5(4)}{3+5} = \frac{-3 + 20}{8} = \frac{17}{8}$.
$z_D = \frac{3(5) + 5(6)}{3+5} = \frac{15 + 30}{8} = \frac{45}{8}$.
Thus,the coordinates of point $D$ are $\left(\frac{22}{8}, \frac{17}{8}, \frac{45}{8}\right)$.

(B) The coordinates of $P_1$ dividing $AB$ in the ratio $1:2$ are given by the section formula:
$P_1 = \left( \frac{1(9) + 2(3)}{1+2}, \frac{1(8) + 2(2)}{1+2}, \frac{1(-10) + 2(-4)}{1+2} \right) = \left( \frac{15}{3}, \frac{12}{3}, \frac{-18}{3} \right) = (5, 4, -6)$.
The coordinates of $P_2$ dividing $AB$ in the ratio $2:1$ are:
$P_2 = \left( \frac{2(9) + 1(3)}{2+1}, \frac{2(8) + 1(2)}{2+1}, \frac{2(-10) + 1(-4)}{2+1} \right) = \left( \frac{21}{3}, \frac{18}{3}, \frac{-24}{3} \right) = (7, 6, -8)$.
Since $P(\alpha, \beta, \gamma)$ is the midpoint of $P_1 P_2$ (ratio $1:1$):
$P = \left( \frac{5+7}{2}, \frac{4+6}{2}, \frac{-6-8}{2} \right) = (6, 5, -7)$.
Thus,$\alpha = 6$,$\beta = 5$,and $\gamma = -7$.
Calculating the expression:
$\alpha + 2\beta + 2\gamma = 6 + 2(5) + 2(-7) = 6 + 10 - 14 = 2$.

(D) Let $B$ divide the line segment $\overline{AC}$ in the ratio $k: 1$. Using the section formula,the coordinates of $B$ are given by:
$B = \left( \frac{k(9) + 1(3)}{k+1}, \frac{k(8) + 1(2)}{k+1}, \frac{k(10) + 1(4)}{k+1} \right)$
Equating the $x$-coordinate:
$\frac{9k + 3}{k+1} = \frac{33}{5}$
$5(9k + 3) = 33(k + 1)$
$45k + 15 = 33k + 33$
$12k = 18$
$k = \frac{18}{12} = \frac{3}{2}$
Thus,the ratio $k: 1$ is $\frac{3}{2}: 1$,which is $3: 2$.

(B) Let $P$ divide the line segment joining $Q(2,2,1)$ and $R(5,1,-2)$ in the ratio $m:1$.
Using the section formula,the coordinates of $P$ are given by:
$P = \left( \frac{5m+2}{m+1}, \frac{m+2}{m+1}, \frac{-2m+1}{m+1} \right)$.
Given that the $x$-coordinate of $P$ is $4$,we have:
$\frac{5m+2}{m+1} = 4$.
Multiplying both sides by $(m+1)$,we get:
$5m + 2 = 4(m + 1) \Rightarrow 5m + 2 = 4m + 4$.
Solving for $m$,we get $m = 2$.
Now,substitute $m = 2$ into the expression for the $z$-coordinate:
$z = \frac{-2(2) + 1}{2 + 1} = \frac{-4 + 1}{3} = \frac{-3}{3} = -1$.
Therefore,the $z$-coordinate of $P$ is $-1$.

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

(D) Let the vertices of the tetrahedron be $A(1,2,3), B(2,-3,1), C(3,2,-1)$ and $D(a, b, c)$.
Since $G\left(\frac{5}{2}, \frac{3}{2}, \frac{9}{4}\right)$ is the centroid of the tetrahedron $ABCD$,we have:
$G = \frac{A+B+C+D}{4}$
$\left(\frac{5}{2}, \frac{3}{2}, \frac{9}{4}\right) = \frac{(1+2+3+a, 2-3+2+b, 3+1-1+c)}{4}$
$\left(\frac{5}{2}, \frac{3}{2}, \frac{9}{4}\right) = \frac{(6+a, 1+b, 3+c)}{4}$
Multiplying by $4$,we get:
$(10, 6, 9) = (6+a, 1+b, 3+c)$
Comparing the coordinates:
$6+a = 10 \Rightarrow a = 4$
$1+b = 6 \Rightarrow b = 5$
$3+c = 9 \Rightarrow c = 6$
Thus,$D = (4, 5, 6)$.
Now,we find the point $P$ that divides $GD$ in the ratio $1:2$ using the section formula:
$P = \left(\frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}, \frac{m z_2 + n z_1}{m+n}\right)$
Here $m=1, n=2$,$G = \left(\frac{5}{2}, \frac{3}{2}, \frac{9}{4}\right)$ and $D = (4, 5, 6)$.
$x = \frac{1(4) + 2(5/2)}{1+2} = \frac{4+5}{3} = \frac{9}{3} = 3$
$y = \frac{1(5) + 2(3/2)}{1+2} = \frac{5+3}{3} = \frac{8}{3}$
$z = \frac{1(6) + 2(9/4)}{1+2} = \frac{6+9/2}{3} = \frac{21/2}{3} = \frac{7}{2}$
Therefore,the required point is $\left(3, \frac{8}{3}, \frac{7}{2}\right)$.

(C) Let the $XOZ$-plane divide the line segment joining the points $A(2, 3, 1)$ and $B(6, 7, 1)$ in the ratio $m: n$.
Using the section formula,the coordinates of the point of division are given by:
$\left(\frac{m(6)+n(2)}{m+n}, \frac{m(7)+n(3)}{m+n}, \frac{m(1)+n(1)}{m+n}\right)$
Since the point lies on the $XOZ$-plane,its $y$-coordinate must be zero.
Therefore,$\frac{7m + 3n}{m+n} = 0$.
$7m + 3n = 0$
$7m = -3n$
$\frac{m}{n} = -\frac{3}{7}$
Thus,the ratio is $-3: 7$.