Question

The line $\frac{x+3}{3}=\frac{y-2}{-2}=\frac{z+1}{1}$ and the plane $4x+5y+3z-5=0$ intersect at a point

• A. $(3,1,-2)$
• B. $(3,-2,1)$
• C. $(2,-1,3)$
• D. $(-1,-2,-3)$

Answer 1

The correct option is B $(3,-2,1)$

Let $\frac{x+3}{3}=\frac{y-2}{-2}=\frac{z+1}{1}=k$

On solving, we get

$\Rightarrow x=3k-3$

$\Rightarrow y=-2k+2$

$\Rightarrow z=k-1$

On substituing these values in the given plane equation we get,

$\Rightarrow 4x+5y+3z-5=0$

$\Rightarrow 4(3k-3)+5(-2k+2)+3(k-1)-5=0$

On simpliying we get,

$\Rightarrow 5k=10$

$\Rightarrow k=2$

Substituting this value of $k$ in equations of $x,y,z$ we get

$\Rightarrow x=3,y=-2,z=1$

Hence point of intersection is $(3,-2,1)$