Question

Find the plane passing through $(4,-1,2)$ and parallel to the lines $\frac{x+2}{3}=\frac{y-2}{-1}=\frac{z+1}{2}$ and $\frac{x-2}{1}=\frac{y-3}{2}=\frac{z-4}{3}$.

• A. $x+y-z-1=0$
• B. $x+y+z+2=0$
• C. $2x+y-z-2=0$
• D. None of these

Answer 1

Answer: (A)

$x+y-z-1=0$

The equation of a plane passing through $(4,-1,2)$ is

$a(x-4)+b(y+1)+c(z-2)=0$

It is parallel to the lines $\frac{x+2}{3}=\frac{y-2}{-1}=\frac{z+1}{2}$ and $\frac{x-2}{1}=\frac{y-3}{2}=\frac{z-4}{3}$

Therefore,

$3a-b+2c=0$

$a+2b+3c=0$

Using cross-multiplication, we get,

$\frac{a}{-3-4}=\frac{b}{2-9}=\frac{c}{6+1}$

$⟹\frac{a}{1}=\frac{b}{1}=\frac{c}{-1}=\lambda \left(say\right)$

$a=\lambda ,b=\lambda ,c=-\lambda$

Substituting the values of $a,b,c$ in equation 1, we get,

$(x-4)+(y+1)-(z-2)=0$

or, $x+y-z-1=0$ as the equation of the required plane.