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The position vectors of three points are $$2\overrightarrow { a } -\overrightarrow { b } +3\overrightarrow { c } ,\overrightarrow { a } -2\overrightarrow { b } +\lambda \overrightarrow { c } $$ and $$\mu \overrightarrow { a } -5\overrightarrow { b } $$, where $$\overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c } $$ are non-coplanar vectors. The points are coliinear when


A
λ=2,μ=94
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B
λ=94,μ=2
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C
λ=94,μ=2
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D
None of these
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Solution

The correct option is D None of these
Let the points $$A,B$$ and $$C$$ respectively, then 
$$\overrightarrow { AB } =\overrightarrow { OB } -\overrightarrow { OA } \\ =\overrightarrow { a } -2\overrightarrow { b } +\lambda \overrightarrow { c } -\left( 2\overrightarrow { a } -\overrightarrow { b } +3\overrightarrow { c }  \right) \\ =-\overrightarrow { a } -\overrightarrow { b } +\left( \lambda -3 \right) \overrightarrow { c } $$
$$\overrightarrow { AC } =\overrightarrow { OC } -\overrightarrow { OA } \\ =\mu \overrightarrow { a } -5\overrightarrow { b } -\left( 2\overrightarrow { a } -\overrightarrow { b } +3\overrightarrow { c }  \right) \\ =\left( \mu -2 \right) \overrightarrow { a } -4\overrightarrow { b } -3\overrightarrow { c } $$
The points are collinear if $$\overrightarrow { AB } =t\overrightarrow { AC } $$
$$\Rightarrow -1=t\left( \mu -2 \right) ,-1=4t,\lambda -3=-3t$$
$$\displaystyle \Rightarrow -1=t\left( \mu -2 \right) ,-1=4t,\lambda -3=-3t$$ and $$\displaystyle -1=\frac { -1 }{ 4 } \left( \mu -2 \right) ,\lambda -3=\frac { 3 }{ 4 } $$
$$\displaystyle \Rightarrow \mu =6,\lambda =\frac {15 }{ 4 } $$

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