Question

On which of the following lines lies the point of intersection of the line, $\frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}$ and the plane, $x+y+z=2$?

• A. $\frac{x-2}{2}=\frac{y-3}{2}=\frac{z+3}{3}$
• B. $\frac{x-4}{1}=\frac{y-5}{1}=\frac{z-5}{-1}$
• C. $\frac{x-1}{1}=\frac{y-3}{2}=\frac{z+4}{-5}$
• D. $\frac{x+3}{3}=\frac{4-y}{3}=\frac{z+1}{-2}$

Answer 1

The correct option is C $\frac{x-1}{1}=\frac{y-3}{2}=\frac{z+4}{-5}$

General point on the given line is

$x=2\lambda +4$
$y=2\lambda +5$
$z=\lambda +3$

Solving with plane,

$2\lambda +4+2\lambda +5+\lambda +3=2$
$5\lambda +12=2$
$5\lambda =-10$
$\lambda =-2$

$x=2\times -2+4=0$
$y=2\times -2+5=1$
$z=1\times -2+3=1$

Points will satisfy option $C$