Question

Find the coordinates of the point of intersection of the line $\frac{x+1}{1}=\frac{y+3}{3}=\frac{z-2}{-2}$ with the plane $3x+4y+5z=5$

Answer 1

$L:\frac{x+1}{1}=\frac{y+3}{3}=\frac{z-2}{-2}=k(x,y,z)=(k-1,3k-3,-2k+2)$

satisfies plane$=3x+4y+5z=53(k-1)+4(3k-3)+5(-2k+2)=5\Rightarrow k=2(x,y,z)=(1,3,-2)$