A) \[8x+14y+13z+37=0\]
B) \[8x-14y+13z+37=0\]
C) \[8x+14y-13z+37=0\]
D) \[8x+14y+13z-37=0\]
E) \[8x-14y-13z-37=0\]
Correct Answer: A
Solution :
Given equation of lines are\[\frac{x-1}{3}=\frac{y+2}{2}=\frac{z}{-4}\]. and\[\frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2}\] Equation of plane is \[a(x-2)+b(y+1)+c(z+3)=0\] ...(i) Now, given lines are parallel to it. \[\therefore \] \[3a+2b-4c=0\] ...(ii) and \[2a-3b+2c=0\] ...(iii) Elimination of a, b and c gives \[\left| \begin{matrix} x-2 & y+1 & z+3 \\ 3 & 2 & -4 \\ 2 & -3 & 2 \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[[(x-2)(4-12)-(y+1)(6+8)+\]\[(z+3)(-9-4)]=0\] \[\Rightarrow \]\[-8x+16-14y-14-13z-39=0\] \[\Rightarrow \] \[8x+14y+13z=-37\] \[\Rightarrow \] \[8x+14y+13z+37=0\]You need to login to perform this action.
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