\displaystyle \textbf{Question 1: } \text{The vertices of the triangle are }A(5,4,6),\ B(1,-1,3)\text{ and }C(4,3,2).
\displaystyle \text{The internal bisector of }\angle A\text{ meets }BC\text{ at }D.\text{ Find the coordinates of }D\text{ and the length }AD.
\displaystyle \text{Answer:}
\displaystyle AB=\sqrt{(5-1)^2+(4+1)^2+(6-3)^2}
\displaystyle =\sqrt{16+25+9}=\sqrt{50}=5\sqrt{2}.
\displaystyle AC=\sqrt{(5-4)^2+(4-3)^2+(6-2)^2}
\displaystyle =\sqrt{1+1+16}=\sqrt{18}=3\sqrt{2}.
\displaystyle \text{Since }AD\text{ is the internal bisector of }\angle BAC,\text{ by the angle bisector theorem,}
\displaystyle \frac{BD}{DC}=\frac{AB}{AC}=\frac{5\sqrt{2}}{3\sqrt{2}}=\frac{5}{3}.
\displaystyle \therefore D\text{ divides }BC\text{ internally in the ratio }5:3.
\displaystyle \therefore D=\left(\frac{5(4)+3(1)}{5+3},\frac{5(3)+3(-1)}{5+3},\frac{5(2)+3(3)}{5+3}\right).
\displaystyle =\left(\frac{23}{8},\frac{12}{8},\frac{19}{8}\right)
\displaystyle =\left(\frac{23}{8},\frac{3}{2},\frac{19}{8}\right).
\displaystyle AD=\sqrt{\left(5-\frac{23}{8}\right)^2+\left(4-\frac{12}{8}\right)^2+\left(6-\frac{19}{8}\right)^2}
\displaystyle =\sqrt{\left(\frac{17}{8}\right)^2+\left(\frac{20}{8}\right)^2+\left(\frac{29}{8}\right)^2}
\displaystyle =\sqrt{\frac{17^2+20^2+29^2}{8^2}}
\displaystyle =\sqrt{\frac{289+400+841}{64}}
\displaystyle =\frac{\sqrt{1530}}{8}=\frac{3\sqrt{170}}{8}.
\displaystyle \therefore D=\left(\frac{23}{8},\frac{3}{2},\frac{19}{8}\right)\text{ and }AD=\frac{3\sqrt{170}}{8}.
\displaystyle \\

\displaystyle \textbf{Question 2: } \text{A point }C\text{ with }z\text{-coordinate }8\text{ lies on the line segment}
\displaystyle \text{joining the points }A(2,-3,4) \ \text{and }B(8,0,10).\text{ Find its coordinates.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }C\text{ divide }AB\text{ internally in the ratio }m:1.
\displaystyle \therefore \text{The coordinates of }C\text{ are } \left(\frac{8m+2}{m+1},\frac{-3}{m+1},\frac{10m+4}{m+1}\right).
\displaystyle \text{Since the }z\text{-coordinate of }C\text{ is }8,
\displaystyle \frac{10m+4}{m+1}=8.
\displaystyle \Rightarrow 10m+4=8m+8.
\displaystyle \Rightarrow 2m=4.
\displaystyle \Rightarrow m=2.
\displaystyle \therefore \text{The coordinates of }C\text{ are } \left(\frac{8(2)+2}{2+1},\frac{-3}{2+1},\frac{10(2)+4}{2+1}\right).
\displaystyle =\left(\frac{18}{3},\frac{-3}{3},\frac{24}{3}\right)=(6,-1,8).
\displaystyle \therefore \text{The required coordinates of }C\text{ are }(6,-1,8).
\displaystyle \\

\displaystyle \textbf{Question 3: } \text{Show that the three points }A(2,3,4),\ B(-1,2,-3)\text{ and }C(-4,1,-10)
\displaystyle \text{ are collinear} \ \text{and find the ratio in which }C\text{ divides }AB.
\displaystyle \text{Answer:}
\displaystyle \text{Let }C\text{ divide }AB\text{ in the ratio }m:1.
\displaystyle \therefore \text{The coordinates of }C\text{ are } \left(\frac{-m+2}{m+1},\frac{2m+3}{m+1},\frac{-3m+4}{m+1}\right).
\displaystyle \text{Since }C=(-4,1,-10),
\displaystyle \frac{-m+2}{m+1}=-4,\qquad \frac{2m+3}{m+1}=1,\qquad \frac{-3m+4}{m+1}=-10.
\displaystyle \Rightarrow -m+2=-4m-4,\qquad 2m+3=m+1,\qquad -3m+4=-10m-10.
\displaystyle \Rightarrow 3m=-6,\qquad m=-2,\qquad 7m=-14.
\displaystyle \therefore m=-2\text{ from all the three equations.}
\displaystyle \therefore C\text{ divides }AB\text{ externally in the ratio }2:1.
\displaystyle \therefore \text{The three points }A,\ B\text{ and }C\text{ are collinear.}
\displaystyle \\

\displaystyle \textbf{Question 4: } \text{Find the ratio in which the line joining }(2,4,5)\text{ and }(3,5,4) \\ \text{is divided by the }yz\text{-plane.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(2,4,5)\text{ and }B(3,5,4).
\displaystyle \text{Suppose the }yz\text{-plane divides }AB\text{ at }P\text{ in the ratio }\lambda:1.
\displaystyle \therefore P=\left(\frac{3\lambda+2}{\lambda+1},\frac{5\lambda+4}{\lambda+1},\frac{4\lambda+5}{\lambda+1}\right).
\displaystyle \text{Since }P\text{ lies on the }yz\text{-plane, its }x\text{-coordinate is }0.
\displaystyle \therefore \frac{3\lambda+2}{\lambda+1}=0.
\displaystyle \Rightarrow 3\lambda+2=0.
\displaystyle \Rightarrow \lambda=-\frac{2}{3}.
\displaystyle \therefore \text{The }yz\text{-plane divides }AB\text{ externally in the ratio }2:3.
\displaystyle \\

\displaystyle \textbf{Question 5: } \text{Find the ratio in which the line joining the points }(2,-1,3)\text{ and }(-1,2,1)
\displaystyle \text{is divided by the plane }x+y+z=5.
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(2,-1,3)\text{ and }B(-1,2,1).
\displaystyle \text{Suppose the plane }x+y+z=5\text{ divides the line joining }A\text{ and }B\text{ at }P
\displaystyle \text{in the ratio }\lambda:1.
\displaystyle \therefore P=\left(\frac{-\lambda+2}{\lambda+1},\frac{2\lambda-1}{\lambda+1},\frac{\lambda+3}{\lambda+1}\right).
\displaystyle \text{Since }P\text{ lies on the plane }x+y+z=5,\text{ its coordinates satisfy the equation of the plane.}
\displaystyle \therefore \frac{-\lambda+2}{\lambda+1}+\frac{2\lambda-1}{\lambda+1}+\frac{\lambda+3}{\lambda+1}=5.
\displaystyle \Rightarrow -\lambda+2+2\lambda-1+\lambda+3=5(\lambda+1).
\displaystyle \Rightarrow 2\lambda+4=5\lambda+5.
\displaystyle \Rightarrow -3\lambda=1.
\displaystyle \Rightarrow \lambda=-\frac{1}{3}.
\displaystyle \therefore \text{The plane divides the line joining }A\text{ and }B\text{ externally in the ratio }1:3.
\displaystyle \\

\displaystyle \textbf{Question 6: } \text{If the points }A(3,2,-4),\ B(9,8,-10)\text{ and }C(5,4,-6)\text{ are} \\ \text{collinear, find the ratio in which }C\text{ divides }AB.
\displaystyle \text{Answer:}
\displaystyle \text{Let }C\text{ divide }AB\text{ in the ratio }\lambda:1.
\displaystyle \therefore C=\left(\frac{9\lambda+3}{\lambda+1},\frac{8\lambda+2}{\lambda+1},\frac{-10\lambda-4}{\lambda+1}\right).
\displaystyle \text{Since }C=(5,4,-6),
\displaystyle \frac{9\lambda+3}{\lambda+1}=5,\qquad \frac{8\lambda+2}{\lambda+1}=4,\qquad \frac{-10\lambda-4}{\lambda+1}=-6.
\displaystyle \text{Each of these equations gives }\lambda=\frac{1}{2}.
\displaystyle \therefore \lambda:1=\frac{1}{2}:1=1:2.
\displaystyle \therefore C\text{ divides }AB\text{ internally in the ratio }1:2.
\displaystyle \\

\displaystyle \textbf{Question 7: } \text{The mid-points of the sides of a triangle }ABC\text{ are }(-2,3,5),
\displaystyle (4,-1,7)\text{ and }(6,5,3). \ \text{Find the coordinates of }A,\ B\text{ and }C.
\displaystyle \text{Answer:}
\displaystyle \text{Let }D(-2,3,5),\ E(4,-1,7)\text{ and }F(6,5,3)\text{ be the mid-points of }BC,\ CA\text{ and }AB\text{ respectively.}
\displaystyle \text{Let }A(x_1,y_1,z_1),\ B(x_2,y_2,z_2)\text{ and }C(x_3,y_3,z_3).
\displaystyle \text{Since }D\text{ is the mid-point of }BC,
\displaystyle \frac{x_2+x_3}{2}=-2,\qquad \frac{y_2+y_3}{2}=3,\qquad \frac{z_2+z_3}{2}=5.
\displaystyle \Rightarrow x_2+x_3=-4,\qquad y_2+y_3=6,\qquad z_2+z_3=10.\qquad\text{... ... ... ... ... (i)}
\displaystyle \text{Since }E\text{ is the mid-point of }CA,
\displaystyle \frac{x_3+x_1}{2}=4,\qquad \frac{y_3+y_1}{2}=-1,\qquad \frac{z_3+z_1}{2}=7.
\displaystyle \Rightarrow x_3+x_1=8,\qquad y_3+y_1=-2,\qquad z_3+z_1=14.\qquad\text{... ... ... ... ... (ii)}
\displaystyle \text{Since }F\text{ is the mid-point of }AB,
\displaystyle \frac{x_1+x_2}{2}=6,\qquad \frac{y_1+y_2}{2}=5,\qquad \frac{z_1+z_2}{2}=3.
\displaystyle \Rightarrow x_1+x_2=12,\qquad y_1+y_2=10,\qquad z_1+z_2=6.\qquad\text{... ... ... ... ... (iii)}
\displaystyle \text{Adding the equations involving }x\text{ in (i), (ii) and (iii),}
\displaystyle 2(x_1+x_2+x_3)=-4+8+12=16.
\displaystyle \Rightarrow x_1+x_2+x_3=8.
\displaystyle \therefore x_1=8-(x_2+x_3)=8-(-4)=12,
\displaystyle x_2=8-(x_3+x_1)=8-8=0,
\displaystyle x_3=8-(x_1+x_2)=8-12=-4.
\displaystyle \text{Adding the equations involving }y\text{ in (i), (ii) and (iii),}
\displaystyle 2(y_1+y_2+y_3)=6-2+10=14.
\displaystyle \Rightarrow y_1+y_2+y_3=7.
\displaystyle \therefore y_1=7-(y_2+y_3)=7-6=1,
\displaystyle y_2=7-(y_3+y_1)=7-(-2)=9,
\displaystyle y_3=7-(y_1+y_2)=7-10=-3.
\displaystyle \text{Adding the equations involving }z\text{ in (i), (ii) and (iii),}
\displaystyle 2(z_1+z_2+z_3)=10+14+6=30.
\displaystyle \Rightarrow z_1+z_2+z_3=15.
\displaystyle \therefore z_1=15-(z_2+z_3)=15-10=5,
\displaystyle z_2=15-(z_3+z_1)=15-14=1,
\displaystyle z_3=15-(z_1+z_2)=15-6=9.
\displaystyle \therefore A(12,1,5),\quad B(0,9,1)\quad\text{and}\quad C(-4,-3,9).
\displaystyle \\

\displaystyle \textbf{Question 8: } A(1,2,3),\ B(0,4,1)\text{ and }C(-1,-1,-3)\text{ are the vertices of a triangle }ABC.
\displaystyle \text{Find the point at which the bisector of }\angle BAC\text{ meets }BC.
\displaystyle \text{Answer:}
\displaystyle AB=\sqrt{(0-1)^2+(4-2)^2+(1-3)^2}
\displaystyle =\sqrt{1+4+4}=\sqrt{9}=3.
\displaystyle AC=\sqrt{(-1-1)^2+(-1-2)^2+(-3-3)^2}
\displaystyle =\sqrt{4+9+36}=\sqrt{49}=7.
\displaystyle \text{Let the internal bisector of }\angle BAC\text{ meet }BC\text{ at }D.
\displaystyle \text{By the angle bisector theorem,}
\displaystyle \frac{BD}{DC}=\frac{AB}{AC}=\frac{3}{7}.
\displaystyle \therefore D\text{ divides }BC\text{ internally in the ratio }3:7.
\displaystyle \therefore D=\left(\frac{3(-1)+7(0)}{3+7},\frac{3(-1)+7(4)}{3+7},\frac{3(-3)+7(1)}{3+7}\right).
\displaystyle =\left(-\frac{3}{10},\frac{25}{10},-\frac{2}{10}\right)
\displaystyle =\left(-\frac{3}{10},\frac{5}{2},-\frac{1}{5}\right).
\displaystyle \therefore \text{The angle bisector meets }BC\text{ at }D\left(-\frac{3}{10},\frac{5}{2},-\frac{1}{5}\right).
\displaystyle \\

\displaystyle \textbf{Question 9: } \text{Find the ratios in which the sphere }x^2+y^2+z^2=504
\displaystyle \text{ divides the line joining the points} \ (12,-4,8)\text{ and }(27,-9,18).
\displaystyle \text{Answer:}
\displaystyle \text{Let the sphere meet the line joining }A(12,-4,8)\text{ and }B(27,-9,18)\text{ at }P(x_1,y_1,z_1).
\displaystyle \therefore x_1^2+y_1^2+z_1^2=504.\qquad\text{... ... ... ... ... (i)}
\displaystyle \text{Let }P\text{ divide the line joining }A\text{ and }B\text{ in the ratio }\lambda:1.
\displaystyle \therefore x_1=\frac{27\lambda+12}{\lambda+1},\qquad y_1=\frac{-9\lambda-4}{\lambda+1},\qquad z_1=\frac{18\lambda+8}{\lambda+1}.
\displaystyle \text{Substituting these values in (i),}
\displaystyle \left(\frac{27\lambda+12}{\lambda+1}\right)^2+\left(\frac{-9\lambda-4}{\lambda+1}\right)^2+\left(\frac{18\lambda+8}{\lambda+1}\right)^2=504.
\displaystyle \Rightarrow \frac{9(9\lambda+4)^2+(9\lambda+4)^2+4(9\lambda+4)^2}{(\lambda+1)^2}=504.
\displaystyle \Rightarrow 14(9\lambda+4)^2=504(\lambda+1)^2.
\displaystyle \Rightarrow (9\lambda+4)^2=36(\lambda+1)^2.
\displaystyle \Rightarrow 9\lambda+4=\pm6(\lambda+1).
\displaystyle \text{Case I: }9\lambda+4=6(\lambda+1)
\displaystyle \Rightarrow 9\lambda+4=6\lambda+6
\displaystyle \Rightarrow 3\lambda=2
\displaystyle \Rightarrow \lambda=\frac{2}{3}.
\displaystyle \therefore \lambda:1=\frac{2}{3}:1=2:3.
\displaystyle \text{Thus, one point of intersection divides }AB\text{ internally in the ratio }2:3.
\displaystyle \text{Case II: }9\lambda+4=-6(\lambda+1)
\displaystyle \Rightarrow 9\lambda+4=-6\lambda-6
\displaystyle \Rightarrow 15\lambda=-10
\displaystyle \Rightarrow \lambda=-\frac{2}{3}.
\displaystyle \text{The negative sign shows that the division is external.}
\displaystyle \therefore \text{The other point of intersection divides }AB\text{ externally in the ratio }2:3.
\displaystyle \therefore \text{The required ratios are }2:3\text{ internally and }2:3\text{ externally.}
\displaystyle \\

\displaystyle \textbf{Question 10: } \text{Show that the plane }ax+by+cz+d=0\text{ divides the line joining the points }
\displaystyle (x_1,y_1,z_1) \ \text{and }(x_2,y_2,z_2)\text{ in the ratio }-\frac{ax_1+by_1+cz_1+d}{ax_2+by_2+cz_2+d}:1.
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(x_1,y_1,z_1)\text{ and }B(x_2,y_2,z_2).
\displaystyle \text{Suppose the plane }ax+by+cz+d=0\text{ divides the line joining }A\text{ and }B\text{ at }P
\displaystyle \text{in the ratio }\lambda:1.
\displaystyle \therefore P=\left(\frac{\lambda x_2+x_1}{\lambda+1},\frac{\lambda y_2+y_1}{\lambda+1},\frac{\lambda z_2+z_1}{\lambda+1}\right).
\displaystyle \text{Since }P\text{ lies on the plane }ax+by+cz+d=0,\text{ its coordinates satisfy the equation of the plane.}
\displaystyle a\left(\frac{\lambda x_2+x_1}{\lambda+1}\right)+b\left(\frac{\lambda y_2+y_1}{\lambda+1}\right)
\displaystyle +c\left(\frac{\lambda z_2+z_1}{\lambda+1}\right)+d=0.
\displaystyle \Rightarrow a(\lambda x_2+x_1)+b(\lambda y_2+y_1)+c(\lambda z_2+z_1)+d(\lambda+1)=0.
\displaystyle \Rightarrow \lambda(ax_2+by_2+cz_2+d)+(ax_1+by_1+cz_1+d)=0.
\displaystyle \Rightarrow \lambda(ax_2+by_2+cz_2+d)=-(ax_1+by_1+cz_1+d).
\displaystyle \Rightarrow \lambda=-\frac{ax_1+by_1+cz_1+d}{ax_2+by_2+cz_2+d}.
\displaystyle \therefore \text{The plane divides the line joining }A\text{ and }B\text{ in the ratio}
\displaystyle -\frac{ax_1+by_1+cz_1+d}{ax_2+by_2+cz_2+d}:1.
\displaystyle \\

\displaystyle \textbf{Question 11: } \text{Find the centroid of a triangle, the midpoints of whose sides are} \\ (1,2,-3),\ (3,0,1)\text{ and }(-1,1,-4).
\displaystyle \text{Answer:}
\displaystyle \text{Let }A(x_1,y_1,z_1),\ B(x_2,y_2,z_2)\text{ and }C(x_3,y_3,z_3)\text{ be the vertices of the triangle.}
\displaystyle \text{Let }D(1,2,-3),\ E(3,0,1)\text{ and }F(-1,1,-4)\text{ be the midpoints of }BC,\ CA\text{ and }AB\text{ respectively.}
\displaystyle D\text{ is the midpoint of }BC.
\displaystyle \therefore \frac{x_2+x_3}{2}=1,\qquad \frac{y_2+y_3}{2}=2,\qquad \frac{z_2+z_3}{2}=-3.
\displaystyle \Rightarrow x_2+x_3=2,\qquad y_2+y_3=4,\qquad z_2+z_3=-6.\qquad\text{... ... ... ... ... (i)}
\displaystyle E\text{ is the midpoint of }CA.
\displaystyle \therefore \frac{x_1+x_3}{2}=3,\qquad \frac{y_1+y_3}{2}=0,\qquad \frac{z_1+z_3}{2}=1.
\displaystyle \Rightarrow x_1+x_3=6,\qquad y_1+y_3=0,\qquad z_1+z_3=2.\qquad\text{... ... ... ... ... (ii)}
\displaystyle F\text{ is the midpoint of }AB.
\displaystyle \therefore \frac{x_1+x_2}{2}=-1,\qquad \frac{y_1+y_2}{2}=1,\qquad \frac{z_1+z_2}{2}=-4.
\displaystyle \Rightarrow x_1+x_2=-2,\qquad y_1+y_2=2,\qquad z_1+z_2=-8.\qquad\text{... ... ... ... ... (iii)}
\displaystyle \text{Adding (i), (ii) and (iii),}
\displaystyle 2(x_1+x_2+x_3)=6\Rightarrow x_1+x_2+x_3=3.
\displaystyle 2(y_1+y_2+y_3)=6\Rightarrow y_1+y_2+y_3=3.
\displaystyle 2(z_1+z_2+z_3)=-12\Rightarrow z_1+z_2+z_3=-6.
\displaystyle \therefore \text{The coordinates of the centroid are}
\displaystyle \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)
\displaystyle =\left(\frac{3}{3},\frac{3}{3},\frac{-6}{3}\right)=(1,1,-2).
\displaystyle \\

\displaystyle \textbf{Question 12: } \text{The centroid of a triangle }ABC\text{ is }(1,1,1). \text{ If the coordinates of }
\displaystyle A\text{ and }B\text{ are }(3,-5,7) \ \text{and }(-1,7,-6)\text{ respectively, find the coordinates of }C.
\displaystyle \text{Answer:}
\displaystyle \text{Let }G\text{ be the centroid of }\triangle ABC.
\displaystyle \text{Given }G=(1,1,1).
\displaystyle \text{Also, }A(3,-5,7)\text{ and }B(-1,7,-6).
\displaystyle \text{Let }C=(x,y,z).
\displaystyle \text{Since the coordinates of the centroid are the averages of the coordinates of the vertices,}
\displaystyle 1=\frac{3+(-1)+x}{3}\Rightarrow \frac{2+x}{3}=1\Rightarrow x=1.
\displaystyle 1=\frac{-5+7+y}{3}\Rightarrow \frac{2+y}{3}=1\Rightarrow y=1.
\displaystyle 1=\frac{7+(-6)+z}{3}\Rightarrow \frac{1+z}{3}=1\Rightarrow z=2.
\displaystyle \therefore C=(1,1,2).
\displaystyle \\

\displaystyle \textbf{Question 13: } \text{Find the coordinates of the points which trisect the line segment} \\ \text{joining the points }P(4,2,-6)\text{ and }Q(10,-16,6).
\displaystyle \text{Answer:}
\displaystyle \text{Given }P(4,2,-6)\text{ and }Q(10,-16,6).
\displaystyle \text{Let }A\text{ and }B\text{ be the points which trisect }PQ.
\displaystyle \therefore PA=AB=BQ.
\displaystyle \therefore PA:AQ=1:2.
\displaystyle \text{Hence }A\text{ divides }PQ\text{ internally in the ratio }1:2.
\displaystyle \therefore A=\left(\frac{1(10)+2(4)}{1+2},\frac{1(-16)+2(2)}{1+2},\frac{1(6)+2(-6)}{1+2}\right).
\displaystyle =\left(\frac{10+8}{3},\frac{-16+4}{3},\frac{6-12}{3}\right).
\displaystyle =\left(\frac{18}{3},\frac{-12}{3},\frac{-6}{3}\right).
\displaystyle =(6,-4,-2).
\displaystyle \text{Also, }PB:BQ=2:1.
\displaystyle \text{Hence }B\text{ divides }PQ\text{ internally in the ratio }2:1.
\displaystyle \therefore B=\left(\frac{2(10)+1(4)}{2+1},\frac{2(-16)+1(2)}{2+1},\frac{2(6)+1(-6)}{2+1}\right).
\displaystyle =\left(\frac{20+4}{3},\frac{-32+2}{3},\frac{12-6}{3}\right).
\displaystyle =\left(\frac{24}{3},\frac{-30}{3},\frac{6}{3}\right).
\displaystyle =(8,-10,2).
\displaystyle \therefore \text{The points of trisection are }(6,-4,-2)\text{ and }(8,-10,2).
\displaystyle \\

\displaystyle \textbf{Question 14: } \text{Using the section formula, show that the points }A(2,-3,4),\ B(-1,2,1) \\ \text{ and }C\left(0,\frac{1}{3},2\right)\text{ are collinear.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }A(2,-3,4),\ B(-1,2,1)\text{ and }C\left(0,\frac{1}{3},2\right).
\displaystyle \text{Let }C\text{ divide }AB\text{ in the ratio }\lambda:1.
\displaystyle \therefore \text{The coordinates of }C\text{ are } \left(\frac{-\lambda+2}{\lambda+1},\frac{2\lambda-3}{\lambda+1},\frac{\lambda+4}{\lambda+1}\right).
\displaystyle \text{Comparing these with the coordinates of }C\left(0,\frac{1}{3},2\right),
\displaystyle \frac{-\lambda+2}{\lambda+1}=0\Rightarrow -\lambda+2=0\Rightarrow \lambda=2.
\displaystyle \frac{2\lambda-3}{\lambda+1}=\frac{1}{3}\Rightarrow 6\lambda-9=\lambda+1\Rightarrow \lambda=2.
\displaystyle \frac{\lambda+4}{\lambda+1}=2\Rightarrow \lambda+4=2\lambda+2\Rightarrow \lambda=2.
\displaystyle \text{Since the same value of }\lambda\text{ is obtained in all the three cases, }C\text{ lies on the line joining }A\text{ and }B.
\displaystyle \therefore \text{The points }A,\ B\text{ and }C\text{ are collinear.}
\displaystyle \\

\displaystyle \textbf{Question 15: } \text{Given that }P(3,2,-4),\ Q(5,4,-6)\text{ and }R(9,8,-10)\text{ are} \\ \text{collinear. Find the ratio in which }Q\text{ divides }PR.
\displaystyle \text{Answer:}
\displaystyle \text{Given }P(3,2,-4),\ Q(5,4,-6)\text{ and }R(9,8,-10)\text{ are collinear.}
\displaystyle \text{Let }Q\text{ divide }PR\text{ in the ratio }\lambda:1.
\displaystyle \therefore \text{The coordinates of }Q\text{ are } \left(\frac{9\lambda+3}{\lambda+1},\frac{8\lambda+2}{\lambda+1},\frac{-10\lambda-4}{\lambda+1}\right).
\displaystyle \text{Comparing these with the coordinates of }Q(5,4,-6),
\displaystyle \frac{9\lambda+3}{\lambda+1}=5\Rightarrow 9\lambda+3=5\lambda+5\Rightarrow \lambda=\frac{1}{2}.
\displaystyle \frac{8\lambda+2}{\lambda+1}=4\Rightarrow 8\lambda+2=4\lambda+4\Rightarrow \lambda=\frac{1}{2}.
\displaystyle \frac{-10\lambda-4}{\lambda+1}=-6\Rightarrow -10\lambda-4=-6\lambda-6\Rightarrow \lambda=\frac{1}{2}.
\displaystyle \text{Since the same value of }\lambda\text{ is obtained in all the three cases,}
\displaystyle Q\text{ divides }PR\text{ in the ratio }\lambda:1=\frac{1}{2}:1=1:2.
\displaystyle \therefore Q\text{ divides }PR\text{ internally in the ratio }1:2.
\displaystyle \\

\displaystyle \textbf{Question 16: } \text{Find the ratio in which the line segment joining the} \\ \text{points }(4,8,10)\text{ and }(6,10,-8)\text{ is divided by the }yz\text{-plane.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }A(4,8,10)\text{ and }B(6,10,-8).
\displaystyle \text{Let the }yz\text{-plane divide }AB\text{ at }P\text{ in the ratio }\lambda:1.
\displaystyle \therefore P=\left(\frac{6\lambda+4}{\lambda+1},\frac{10\lambda+8}{\lambda+1},\frac{-8\lambda+10}{\lambda+1}\right).
\displaystyle \text{Since }P\text{ lies on the }yz\text{-plane, its }x\text{-coordinate is }0.
\displaystyle \therefore \frac{6\lambda+4}{\lambda+1}=0.
\displaystyle \Rightarrow 6\lambda+4=0.
\displaystyle \Rightarrow \lambda=-\frac{2}{3}.
\displaystyle \text{Since }\lambda\text{ is negative, the division is external.}
\displaystyle \therefore \text{The }yz\text{-plane divides }AB\text{ externally in the ratio }2:3.
\displaystyle \\

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