Question

The point of intersection of the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z+2}{3}$ and the plane $2x+3y+z=0$ is:

• A. (a) (0, 1, -2)
• B. (b) (1, 2, 3)
• C. (c) (-1, 9, -25)
• D. (d) $\left(\frac{-1}{11},\frac{9}{11},\frac{-25}{11}\right)$

Answer 1

Answer: (D)

(d) $\left(\frac{-1}{11},\frac{9}{11},\frac{-25}{11}\right)$

$\frac{x}{1}=\frac{y-1}{2}=\frac{z+2}{3}=\lambda 2x+3y+z=0$

coordinates of line

$x=\lambda ,y=2\lambda +1,z=3\lambda -2$

substituting these coordinates in the equation of plane

$2\left(\lambda \right)+3(2\lambda +1)+(3\lambda -2)=0$

$\Rightarrow 2\lambda +\lambda +3+3\lambda -2=0$

$\Rightarrow 11\lambda +1=0\Rightarrow \lambda =\frac{-1}{11}$

Intersection: $x=\frac{-1}{11},y=2\left(\frac{-1}{11}\right)+1,z=3\left(\frac{-1}{11}\right)-2$

$\Rightarrow x=\frac{-1}{11},y=\frac{9}{11},z=\frac{-25}{11}$

The point of intersection is $(\frac{-1}{11},\frac{9}{11},\frac{-25}{11})$