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  • Let the lines <img alt="\mathrm{\frac{x-1}{\lambda}=\frac{y-2}{1}=\frac{z-3}{2}}" src="/latex-image/?%5Cmathrm%7B%5Cfrac%7Bx-1%7D%7B%5Clambda%7D%3D%5Cfrac%7By-2%7D%7B1%7D%3D%5Cfrac%7Bz-3%7D%7B

Let the lines \mathrm{\frac{x-1}{\lambda}=\frac{y-2}{1}=\frac{z-3}{2}} and \mathrm{\frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda}} be coplanar and \mathrm{p} be the plane containing these two lines. Then which of the following points does NOT lie on \mathrm{p} ?

Option: 1

(0,-2,-2)


Option: 2

(-5,0,-1)


Option: 3

(3,-1,0)


Option: 4

(0,4,5)


Answers (1)

best_answer

\begin{aligned} &\text{Using condition for coplanarity}\\ &\left|\begin{array}{ccc}\mathrm{1-(-26)} & \mathrm{2-(-18) }& \mathrm{3-(-28)} \\ \mathrm{2} &\mathrm{ 1} &\mathrm{ 2 }\\ \mathrm{-2 }& \mathrm{3} & \mathrm{2}\end{array}\right|=0\\ &\Rightarrow \mathrm{\quad r_1=3}\\ &\text{Equation of plane}\\ &\left|\begin{array}{ccc}\mathrm{x-1 }& \mathrm{y-2} & \mathrm{z-3} \\ \mathrm{3} &\mathrm{1 }& \mathrm{2 }\\\mathrm{ -2} & \mathrm{3} & \mathrm{3}\end{array}\right|=0\\ &\Rightarrow \quad 3 x+13 y-11 z+4=0\\ &\text{Checking options, option (D) does not satisfy it}.\\ &\therefore \text{option (D)} \end{aligned}

 

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Rishabh

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