"Find the parametric equations of a line that passes through point $(1,1,0)$, parallel to plane $2x+3y+z=7$ and perpendicular to the line $\frac{x-1}{-2}= \frac{y}{3}=-z-2$"
I don't know where to start, I know how to find the equations when given two different points in space.
The equation of the line is $x=1+at$, $y=1+bt$, $z=ct$. The vector $\vec{d}=(a,b,c)$ is parallel to the line, so it must be orthogonal to $\vec{n}=(2,3,1)$. So $2a+3b+c=0$. A parallel vector to the line $\frac{x-1}{-2}= \frac{y}{3}=-z-2$ is $(-2,3,-1)$, so $(a,b,c)\cdot (-2,3,-1)=0$. Hence $-2a+3b-c=0$. So combining this with $2a+3b+c=0$, and adding , we get $6b=0$ so $b=0$.We also get $2a=c$. We can let $c=1$ and $a=2$. and $x=1+2t,y=1,z=t$. I hope I made no mistake!