Question

The equation of the plane through the point $ (2,-1,-3) $ and parallel to the lines $ \frac{x-1}{3}=\frac{y+2}{2}=\frac{z}{-4} $ and $ \frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2} $ is:

• 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 $

Answer 1

Answer: (A)

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 $