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 $\left|\begin{array}{ccc}x-2& y+1& z+3\\ 3& 2& -4\\ 2& -3& 2\end{array}\right|=0$ $\Rightarrow$ $[(x-2)(4-12)-(y+1)(6+8)+$ $(z+3)(-9-4)]=0$
$\Rightarrow$ $-8x+16-14y-14-13z-39=0$
$\Rightarrow$ $8x+14y+13z=-37$
$\Rightarrow$ $8x+14y+13z+37=0$