\displaystyle \textbf{Question 1: }\text{Show that the points whose position vectors are as given below are } \\ \text{collinear:}\\ \text{(i) }2\hat{i}+\hat{j}-\hat{k},\ 3\hat{i}-2\hat{j}+\hat{k}\text{ and }\hat{i}+4\hat{j}-3\hat{k}\\ \text{(ii) }3\hat{i}-2\hat{j}+4\hat{k},\ \hat{i}+\hat{j}+\hat{k}\text{ and }-\hat{i}+4\hat{j}-2\hat{k}.
\displaystyle \text{Answer:}
\displaystyle \text{(i) }
\displaystyle \text{Let the points be }A,B,C\text{ with position vectors }2\widehat{i}+\widehat{j}-\widehat{k},\;3\widehat{i}-2\widehat{j}+\widehat{k}\text{ and }\widehat{i}+4\widehat{j}-3\widehat{k}.
\displaystyle \text{Then, }\overrightarrow{AB}=\text{Position vector of }B-\text{Position vector of }A
\displaystyle =(3\widehat{i}-2\widehat{j}+\widehat{k})-(2\widehat{i}+\widehat{j}-\widehat{k})
\displaystyle =\widehat{i}-3\widehat{j}+2\widehat{k}
\displaystyle \overrightarrow{BC}=\text{Position vector of }C-\text{Position vector of }B
\displaystyle =(\widehat{i}+4\widehat{j}-3\widehat{k})-(3\widehat{i}-2\widehat{j}+\widehat{k})
\displaystyle =-2\widehat{i}+6\widehat{j}-4\widehat{k}
\displaystyle =-2(\widehat{i}-3\widehat{j}+2\widehat{k})
\displaystyle \therefore \overrightarrow{BC}=-2\overrightarrow{AB}
\displaystyle \text{So, }\overrightarrow{AB}\text{ and }\overrightarrow{BC}\text{ are parallel vectors. But }B\text{ is a common point to them.}
\displaystyle \text{Hence, }A,B\text{ and }C\text{ are collinear.}
\displaystyle \text{(ii) }
\displaystyle \text{Let the points be }A,B,C\text{ with position vectors }3\widehat{i}-2\widehat{j}+4\widehat{k},\;\widehat{i}+\widehat{j}+\widehat{k}\text{ and }-\widehat{i}+4\widehat{j}-2\widehat{k}\text{ respectively.}
\displaystyle \text{Then, }\overrightarrow{AB}=\text{Position vector of }B-\text{Position vector of }A
\displaystyle =(\widehat{i}+\widehat{j}+\widehat{k})-(3\widehat{i}-2\widehat{j}+4\widehat{k})
\displaystyle =-2\widehat{i}+3\widehat{j}-3\widehat{k}
\displaystyle \overrightarrow{BC}=\text{Position vector of }C-\text{Position vector of }B
\displaystyle =(-\widehat{i}+4\widehat{j}-2\widehat{k})-(\widehat{i}+\widehat{j}+\widehat{k})
\displaystyle =-2\widehat{i}+3\widehat{j}-3\widehat{k}
\displaystyle \therefore \overrightarrow{AB}=\overrightarrow{BC}
\displaystyle \text{So, }\overrightarrow{AB}\text{ and }\overrightarrow{BC}\text{ are parallel vectors. But }B\text{ is a common point to them.}
\displaystyle \text{Hence, }A,B\text{ and }C\text{ are collinear.}

\displaystyle \textbf{Question 2: }\text{Using vector method, prove that the following points are collinear:}\\ \text{(i) }A(6,-7,-1),\ B(2,-3,1)\text{ and }C(4,-5,0)\\ \text{(ii) }A(2,-1,3),\ B(4,3,1)\text{ and }C(3,1,2)\\ \text{(iii) }A(1,2,7),\ B(2,6,3)\text{ and }C(3,10,-1)\\ \text{(iv) }A(-3,-2,-5),\ B(1,2,3)\text{ and }C(3,4,7).
\displaystyle \text{Answer:}
\displaystyle \text{(i) }
\displaystyle \text{Given the points }A(6,-7,-1),\;B(2,-3,1)\text{ and }C(4,-5,0).
\displaystyle \text{Then, }\overrightarrow{AB}=\text{Position vector of }B-\text{Position vector of }A
\displaystyle =(2\widehat{i}-3\widehat{j}+\widehat{k})-(6\widehat{i}-7\widehat{j}-\widehat{k})
\displaystyle =-4\widehat{i}+4\widehat{j}+2\widehat{k}
\displaystyle =-2(2\widehat{i}-2\widehat{j}-\widehat{k})
\displaystyle \overrightarrow{BC}=\text{Position vector of }C-\text{Position vector of }B
\displaystyle =(4\widehat{i}-5\widehat{j}+0\widehat{k})-(2\widehat{i}-3\widehat{j}+\widehat{k})
\displaystyle =2\widehat{i}-2\widehat{j}-\widehat{k}
\displaystyle \therefore \overrightarrow{AB}=-2\overrightarrow{BC}
\displaystyle \text{So, }\overrightarrow{AB}\text{ and }\overrightarrow{BC}\text{ are parallel vectors. But }B\text{ is a common point to them.}
\displaystyle \text{Hence, the given points }A
\displaystyle \text{(ii) }
\displaystyle \text{Given the points }A(2,-1,3),\;B(4,3,1)\text{ and }C(3,1,2).
\displaystyle \text{Then, }\overrightarrow{AB}=\text{Position vector of }B-\text{Position vector of }A
\displaystyle =(4\widehat{i}+3\widehat{j}+\widehat{k})-(2\widehat{i}-\widehat{j}+3\widehat{k})
\displaystyle =2\widehat{i}+4\widehat{j}-2\widehat{k}
\displaystyle =-2(-\widehat{i}-2\widehat{j}+\widehat{k})
\displaystyle \overrightarrow{BC}=\text{Position vector of }C-\text{Position vector of }B
\displaystyle =(3\widehat{i}+\widehat{j}+2\widehat{k})-(4\widehat{i}+3\widehat{j}+\widehat{k})
\displaystyle =-\widehat{i}-2\widehat{j}+\widehat{k}
\displaystyle \therefore \overrightarrow{AB}=-2\overrightarrow{BC}
\displaystyle \text{So, }\overrightarrow{AB}\text{ and }\overrightarrow{BC}\text{ are parallel vectors. But }B\text{ is a common point to them.}
\displaystyle \text{Hence, the given points }A,B\text{ and }C\text{ are collinear.}
\displaystyle \text{(iii) }
\displaystyle \text{Given the points }A(1,2,7),\;B(2,6,3)\text{ and }C(3,10,-1).
\displaystyle \text{Then, }\overrightarrow{AB}=\text{Position vector of }B-\text{Position vector of }A
\displaystyle =(2\widehat{i}+6\widehat{j}+3\widehat{k})-(\widehat{i}+2\widehat{j}+7\widehat{k})
\displaystyle =\widehat{i}+4\widehat{j}-4\widehat{k}
\displaystyle \overrightarrow{BC}=\text{Position vector of }C-\text{Position vector of }B
\displaystyle =(3\widehat{i}+10\widehat{j}-\widehat{k})-(2\widehat{i}+6\widehat{j}+3\widehat{k})
\displaystyle =\widehat{i}+4\widehat{j}-4\widehat{k}
\displaystyle \therefore \overrightarrow{AB}=\overrightarrow{BC}
\displaystyle \text{So, }\overrightarrow{AB}\text{ and }\overrightarrow{BC}\text{ are parallel vectors. But }B\text{ is a common point to them.}
\displaystyle \text{Hence, the given points }A,B\text{ and }C\text{ are collinear.}
\displaystyle \text{(iv) }
\displaystyle \text{Given the points }A(-3,-2,-5),\;B(1,2,3)\text{ and }C(3,4,7).
\displaystyle \overrightarrow{AB}=\text{Position vector of }B-\text{Position vector of }A
\displaystyle =( \widehat{i}+2\widehat{j}+3\widehat{k})-(-3\widehat{i}-2\widehat{j}-5\widehat{k})
\displaystyle =4\widehat{i}+4\widehat{j}+8\widehat{k}
\displaystyle =2(2\widehat{i}+2\widehat{j}+4\widehat{k})
\displaystyle \overrightarrow{BC}=\text{Position vector of }C-\text{Position vector of }B
\displaystyle =(3\widehat{i}+4\widehat{j}+7\widehat{k})-(\widehat{i}+2\widehat{j}+3\widehat{k})
\displaystyle =2\widehat{i}+2\widehat{j}+4\widehat{k}
\displaystyle \therefore\ \overrightarrow{AB}=2\overrightarrow{BC}
\displaystyle \text{So, }\overrightarrow{AB}\text{ and }\overrightarrow{BC}\text{ are parallel vectors. But }B\text{ is a common point to them.}
\displaystyle \text{Hence, the given points }A,B\text{ and }C\text{ are collinear.}

\displaystyle \textbf{Question 3: }\text{If }\vec{a},\vec{b},\vec{c}\text{ are non-zero, non-coplanar vectors, prove that the following} \\ \text{vectors are coplanar:}\\ \text{(i) }5\vec{a}+6\vec{b}+7\vec{c},\ 7\vec{a}-8\vec{b}+9\vec{c}\text{ and }3\vec{a}+20\vec{b}+5\vec{c}\\ \text{(ii) }\vec{a}-2\vec{b}+3\vec{c},\ \vec{a}-3\vec{b}+5\vec{c}\text{ and }-2\vec{a}+3\vec{b}-4\vec{c}.
\displaystyle \text{Answer:}
\displaystyle \textbf{(i)}\ \text{The three vectors are coplanar if one of them is expressible as a linear } \\ \text{combination of the other two. Let}
\displaystyle 5\overrightarrow{a}+6\overrightarrow{b}+7\overrightarrow{c}=x(7\overrightarrow{a}-8\overrightarrow{b}+9\overrightarrow{c})+y(3\overrightarrow{a}+20\overrightarrow{b}+5\overrightarrow{c})
\displaystyle =(7x+3y)\overrightarrow{a}+(-8x+20y)\overrightarrow{b}+(9x+5y)\overrightarrow{c}
\displaystyle \Rightarrow 7x+3y=5,\;-8x+20y=6,\;9x+5y=7
\displaystyle \text{Solving the first two equations, we get }x=\frac{1}{2},\;y=\frac{1}{2}
\displaystyle \text{Clearly, these values of }x\text{ and }y\text{ satisfy the third equation also.}
\displaystyle \text{Hence, the given vectors are coplanar.}
\displaystyle \textbf{(ii)}\ \text{The three vectors are coplanar if one of them is expressible as a linear } \\ \text{combination of the other two. Let}
\displaystyle \overrightarrow{a}-2\overrightarrow{b}+3\overrightarrow{c}=x(-3\overrightarrow{b}+5\overrightarrow{c})+y(-2\overrightarrow{a}+3\overrightarrow{b}-4\overrightarrow{c})
\displaystyle =(-2y)\overrightarrow{a}+(-3x+3y)\overrightarrow{b}+(5x-4y)\overrightarrow{c}
\displaystyle \Rightarrow -2y=1,\;-3x+3y=-2,\;5x-4y=3
\displaystyle \text{Solving the first two equations, we get }y=-\frac{1}{2},\;x=\frac{1}{6}
\displaystyle \text{These values of }x\text{ and }y\text{ do not satisfy the third equation.}
\displaystyle \text{Hence, the given vectors are not coplanar.}

\displaystyle \textbf{Question 4: }\text{Show that the four points having position vectors }6\hat{i}-7\hat{j},\ \\ 16\hat{i}-19\hat{j}-4\hat{k},\ 3\hat{j}-6\hat{k},\ 2\hat{i}-5\hat{j}+10\hat{k}\text{ are coplanar.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the given four points be }P,Q,R,S\text{ respectively. Three points are coplanar if the vectors }  \\ \overrightarrow{PQ},\overrightarrow{PR}\text{ and }\overrightarrow{PS}\text{ are coplanar.}
\displaystyle \text{These vectors are coplanar if one of them can be expressed as a linear combination } \\ \text{of the other two. So, let }
\displaystyle \overrightarrow{PQ}=x\overrightarrow{PR}+y\overrightarrow{PS}
\displaystyle 10\widehat{i}-12\widehat{j}-4\widehat{k}=x(-6\widehat{i}+10\widehat{j}-6\widehat{k})+y(-4\widehat{i}+2\widehat{j}+10\widehat{k})
\displaystyle =(-6x-4y)\widehat{i}+(10x+2y)\widehat{j}+(-6x+10y)\widehat{k}
\displaystyle \Rightarrow -6x-4y=10,\;10x+2y=-12,\;-6x+10y=-4
\displaystyle \text{Solving the first two equations, we get }x=-1,\;y=-1
\displaystyle \text{These values also satisfy the third equation.}
\displaystyle \text{Hence, the given four points are coplanar.}

\displaystyle \textbf{Question 5: }\text{Prove that the following vectors are coplanar:}\\ \text{(i) }2\hat{i}-\hat{j}+\hat{k},\ \hat{i}-3\hat{j}-5\hat{k}\text{ and }3\hat{i}-4\hat{j}-4\hat{k}\\ \text{(ii) }\hat{i}+\hat{j}+\hat{k},\ 2\hat{i}+3\hat{j}-\hat{k}\text{ and }-\hat{i}-2\hat{j}+2\hat{k}.
\displaystyle \text{Answer:}
\displaystyle \text{(i) }
\displaystyle \text{Given the vectors }P(2\widehat{i}-\widehat{j}+\widehat{k}),\;Q(\widehat{i}-3\widehat{j}-5\widehat{k})\text{ and }R(3\widehat{i}-4\widehat{j}-4\widehat{k}).
\displaystyle \text{We know that three vectors are coplanar if one of them is expressible as a linear } \\ \text{combination of the other two.}
\displaystyle \text{Let }2\widehat{i}-\widehat{j}+\widehat{k}=x(\widehat{i}-3\widehat{j}-5\widehat{k})+y(3\widehat{i}-4\widehat{j}-4\widehat{k}).
\displaystyle =\widehat{i}(x+3y)+\widehat{j}(-3x-4y)+\widehat{k}(-5x-4y).
\displaystyle \Rightarrow x+3y=2,\;-3x-4y=-1,\;-5x-4y=1.
\displaystyle \text{Solving the first two equations, we get }x=-1,\;y=1.
\displaystyle \text{These values also satisfy the third equation.}
\displaystyle \text{Hence, the given vectors are coplanar.}
\displaystyle \text{(ii) }
\displaystyle \text{Given the vectors }P(\widehat{i}+\widehat{j}+\widehat{k}),\;Q(2\widehat{i}+3\widehat{j}-\widehat{k})\text{ and }R(-\widehat{i}-2\widehat{j}+2\widehat{k}).
\displaystyle \text{We know that three vectors are coplanar if one of them is expressible as a linear } \\ \text{combination of the other two.}
\displaystyle \text{Let }\widehat{i}+\widehat{j}+\widehat{k}=x(2\widehat{i}+3\widehat{j}-\widehat{k})+y(-\widehat{i}-2\widehat{j}+2\widehat{k}).
\displaystyle =\widehat{i}(2x-y)+\widehat{j}(3x-2y)+\widehat{k}(-x+2y).
\displaystyle \Rightarrow 2x-y=1,\;3x-2y=1,\;-x+2y=1.
\displaystyle \text{Solving the first two equations, we get }x=1,\;y=1.
\displaystyle \text{These values also satisfy the third equation.}
\displaystyle \text{Hence, the given vectors are coplanar.}

\displaystyle \textbf{Question 6: }\text{Prove that the following vectors are non-coplanar:}\\ \text{(i) }3\hat{i}+\hat{j}-\hat{k},\ 2\hat{i}-\hat{j}+7\hat{k}\text{ and }7\hat{i}-\hat{j}+23\hat{k}\\ \text{(ii) }\hat{i}+2\hat{j}+3\hat{k},\ 2\hat{i}+\hat{j}+3\hat{k}\text{ and }\hat{i}+\hat{j}+\hat{k}.
\displaystyle \text{Answer:}
\displaystyle \text{(i) }
\displaystyle \text{Let, if possible, the given vectors be coplanar. Then one of the vectors is expressible as a linear} \\ \text{combination of the other two.}
\displaystyle 3\widehat{i}+\widehat{j}-\widehat{k}=x(2\widehat{i}-\widehat{j}+7\widehat{k})+y(7\widehat{i}-\widehat{j}+23\widehat{k}).
\displaystyle =\widehat{i}(2x+7y)+\widehat{j}(-x-y)+\widehat{k}(7x+23y).
\displaystyle \Rightarrow 2x+7y=3,\;-x-y=1,\;7x+23y=-1.
\displaystyle \text{Solving the first two equations, we get }x=-2,\;y=1.
\displaystyle \text{These values do not satisfy the third equation.}
\displaystyle \text{Hence, the given vectors are non-coplanar.}
\displaystyle \text{(ii) }
\displaystyle \text{Let, if possible, the given vectors be coplanar. Then one of the vectors is expressible in} \\ \text{terms of the other two.}
\displaystyle \widehat{i}+2\widehat{j}+3\widehat{k}=x(2\widehat{i}+\widehat{j}+3\widehat{k})+y(\widehat{i}+\widehat{j}+\widehat{k}).
\displaystyle =\widehat{i}(2x+y)+\widehat{j}(x+y)+\widehat{k}(3x+y).
\displaystyle \Rightarrow 2x+y=1,\;x+y=2,\;3x+y=3.
\displaystyle \text{Solving the first two equations, we get }x=-1,\;y=3.
\displaystyle \text{These values do not satisfy the third equation.}
\displaystyle \text{Hence, the given vectors are non-coplanar.}

\displaystyle \textbf{Question 7: }\text{If }\vec{a},\vec{b},\vec{c}\text{ are non-coplanar vectors, prove that the following } \\ \text{vectors are non-coplanar:}\\ \text{(i) }2\vec{a}-\vec{b}+3\vec{c},\ \vec{a}+\vec{b}-2\vec{c}\text{ and }\vec{a}+\vec{b}-3\vec{c}\\ \text{(ii) }\vec{a}+2\vec{b}+3\vec{c},\ 2\vec{a}+\vec{b}+3\vec{c}\text{ and }\vec{a}+\vec{b}+\vec{c}.
\displaystyle \text{Answer:}
\displaystyle \text{(i) }
\displaystyle \text{Let, if possible, the following vectors be coplanar. Then one of the vectors is expressible in} \\ \text{terms of the other two.}
\displaystyle 2\vec{a}-\vec{b}+3\vec{c}=x(\vec{a}+\vec{b}-2\vec{c})+y(\vec{a}+\vec{b}-3\vec{c}).
\displaystyle =\vec{a}(x+y)+\vec{b}(x+y)+\vec{c}(-2x-3y).
\displaystyle \Rightarrow x+y=2,\;x+y=-1,\;-2x-3y=3.
\displaystyle \text{But }x+y=2\neq -1.
\displaystyle \text{Hence, the given vectors are non-coplanar.}
\displaystyle \text{(ii) }
\displaystyle \text{Let, if possible, the following vectors be coplanar. Then one of the vectors is expressible in} \\ \text{terms of the other two.}
\displaystyle \vec{a}+2\vec{b}+3\vec{c}=x(2\vec{a}+\vec{b}+3\vec{c})+y(\vec{a}+\vec{b}+\vec{c}).
\displaystyle =\vec{a}(2x+y)+\vec{b}(x+y)+\vec{c}(3x+y).
\displaystyle \Rightarrow 2x+y=1,\;x+y=2,\;3x+y=3.
\displaystyle \text{Solving the first two equations, we get }x=-1,\;y=3.
\displaystyle \text{These values do not satisfy the third equation.}
\displaystyle \text{Hence, the given vectors are non-coplanar.}

\displaystyle \textbf{Question 8: }\text{Show that the vectors }\vec{a},\vec{b},\vec{c}\text{ given by }\vec{a}=\hat{i}+2\hat{j}+3\hat{k}, \\ \vec{b}=2\hat{i}+\hat{j}+3\hat{k}\text{ and }\vec{c}=\hat{i}+\hat{j}+\hat{k}\text{ are non-coplanar. Express vector } \\ \vec{d}=2\hat{i}-\hat{j}-3\hat{k}\text{ as a linear combination of the vectors }\vec{a},\vec{b}\text{ and }\vec{c}.
\displaystyle \text{Answer:}
\displaystyle \text{Let the given vectors }\vec{a}=\hat{i}+2\hat{j}+3\hat{k},\;\vec{b}=2\hat{i}+\hat{j}+3\hat{k}\text{ and }\vec{c}=\hat{i}+\hat{j}+\hat{k}\text{ be coplanar.}
\displaystyle \text{Then one of the vectors is expressible as a linear combination of the other two.}
\displaystyle \hat{i}+2\hat{j}+3\hat{k}=x(2\hat{i}+\hat{j}+3\hat{k})+y(\hat{i}+\hat{j}+\hat{k}).
\displaystyle =\hat{i}(2x+y)+\hat{j}(x+y)+\hat{k}(3x+y).
\displaystyle \Rightarrow 2x+y=1,\;x+y=2,\;3x+y=3.
\displaystyle \text{Solving the first two equations, we get }x=-1,\;y=3.
\displaystyle \text{These values do not satisfy the third equation.}
\displaystyle \text{Hence, the given vectors are non-coplanar.}
\displaystyle \text{Now, }\vec{d}=2\hat{i}-\hat{j}-3\hat{k}\text{ which can be expressed as}
\displaystyle 2\hat{i}-\hat{j}-3\hat{k}=x(\hat{i}+2\hat{j}+3\hat{k})+y(2\hat{i}+\hat{j}+3\hat{k})+z(\hat{i}+\hat{j}+\hat{k}).
\displaystyle =\hat{i}(x+2y+z)+\hat{j}(2x+y+z)+\hat{k}(3x+3y+z).
\displaystyle \Rightarrow x+2y+z=2,\;2x+y+z=-1,\;3x+3y+z=-3.
\displaystyle \Rightarrow x=-\frac{8}{3},\;y=\frac{1}{3},\;z=4.
\displaystyle \text{Hence, }\vec{d}\text{ is expressible as a linear combination of }\vec{a},\vec{b}\text{ and }\vec{c}.

\displaystyle \textbf{Question 9: }\text{Prove that a necessary and sufficient condition for three vectors } \\ \vec{a},\vec{b} \text{ and }\vec{c}\text{ to be coplanar is that there exist scalars }l,m,n\text{ not all zero } \\ \text{simultaneously such that }l\vec{a}+m\vec{b}+n\vec{c}=\vec{0}.
\displaystyle \text{Answer:}
\displaystyle \textbf{Necessary Condition:}
\displaystyle \text{Let }\vec{a},\vec{b},\vec{c}\text{ be three coplanar vectors. Then one of them is expressible as a linear} \\ \text{combination of the other two.}
\displaystyle \vec{c}=x\vec{a}+y\vec{b}\text{ for some scalars }x,y.
\displaystyle \Rightarrow x\vec{a}+y\vec{b}-\vec{c}=\vec{0}.
\displaystyle \Rightarrow l\vec{a}+m\vec{b}+n\vec{c}=\vec{0},\text{ where }l=x,\;m=y,\;n=-1.
\displaystyle \text{Thus, if }\vec{a},\vec{b},\vec{c}\text{ are coplanar, there exist scalars }l,m,n\text{ (not all zero) such that }l\vec{a}+m\vec{b}+n\vec{c}=\vec{0}.
\displaystyle \textbf{Sufficient Condition:}
\displaystyle \text{Let }\vec{a},\vec{b},\vec{c}\text{ be such that there exist scalars }l,m,n\text{ (not all zero) satisfying }l\vec{a}+m\vec{b}+n\vec{c}=\vec{0}.
\displaystyle \Rightarrow n\vec{c}=-l\vec{a}-m\vec{b}.
\displaystyle \Rightarrow \vec{c}=-\frac{l}{n}\vec{a}-\frac{m}{n}\vec{b}.
\displaystyle \Rightarrow \vec{c}\text{ is a linear combination of }\vec{a}\text{ and }\vec{b}.
\displaystyle \text{Hence, }\vec{a},\vec{b},\vec{c}\text{ are coplanar vectors.}

\displaystyle \textbf{Question 10: }\text{Show that the four points }A,B,C\text{ and }D\text{ with position vectors } \\ \vec{a},\vec{b},\vec{c}\text{ and }\vec{d}\text{ respectively are coplanar if and } \text{only if }3\vec{a}-2\vec{b}+\vec{c}-2\vec{d}=\vec{0}.
\displaystyle \text{Answer:}
\displaystyle \textbf{Necessary Condition:}
\displaystyle \text{Firstly, let }\vec{a},\vec{b},\vec{c}\text{ be coplanar vectors. Then one of them is expressible as} \\ \text{a linear combination of the other two.}
\displaystyle \vec{c}=x\vec{a}+y\vec{b}\text{ for some scalars }x,y.
\displaystyle \Rightarrow x\vec{a}+y\vec{b}-\vec{c}=\vec{0}.
\displaystyle \Rightarrow l\vec{a}+m\vec{b}+n\vec{c}=\vec{0},\text{ where }l=x,\;m=y,\;n=-1.
\displaystyle \text{Thus, if }\vec{a},\vec{b},\vec{c}\text{ are coplanar, there exist scalars }l,m,n\text{ (not all zero) such that }l\vec{a}+m\vec{b}+n\vec{c}=\vec{0}.
\displaystyle \textbf{Sufficient Condition:}
\displaystyle \text{Let }\vec{a},\vec{b},\vec{c}\text{ be such that there exist scalars }l,m,n\text{ (not all zero) satisfying }l\vec{a}+m\vec{b}+n\vec{c}=\vec{0}.
\displaystyle \Rightarrow n\vec{c}=-l\vec{a}-m\vec{b}.
\displaystyle \Rightarrow \vec{c}=-\frac{l}{n}\vec{a}-\frac{m}{n}\vec{b}.
\displaystyle \Rightarrow \vec{c}\text{ is a linear combination of }\vec{a}\text{ and }\vec{b}.
\displaystyle \Rightarrow \vec{c}\text{ lies in the plane of }\vec{a}\text{ and }\vec{b}.
\displaystyle \text{Hence, }\vec{a},\vec{b},\vec{c}\text{ are coplanar vectors.}

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