How do i define the equation for a line with the following information? I have to find the parametric equation for the line $M_1$ with the following info: 


*

*$M_1$ goes through the point $P(1,2,2)$

*Is parallel with the plane $x + 3y + z = 1$

*Has an intersection somewhere with the line $M_2 = 1+t, 2-2t, -1+2t$
So if I understand it correctly, since $M_1$ is parallel with the plane $x + 3y + z = 1$.  I can do $x = 1-3y-z$ and replace $y$ and $z$ with $t$ and $s$ to get the direction of $M_1$. The planes equation will then be:
$$(x,y,z)=(1-3s-t , s , t)$$
So then the $M_1$ should be with the coordinates from $P$:
$$x = 1 + 1-3s-t$$
$$y = 2 + s$$
$$z = 2 + t$$
But it is incorrect because $M_1$ and $M_2$ have no intersections. Where did I go wrong here?
Thanks in advance for the answers!
 A: I expect there is a typo in the expression for $M_2$. You've written $-1+2$ for the third coordinate which seems incorrect.  I'll guess you meant $-1+2t$ and compute from there.  If you meant something else, you should be able to adapt the calculation easily enough.
We note that the normal to the given plane is $$\vec n=(1,3,1)$$
We assume the intersection between the desired line and the given line is $P(t)=(1+t,2-2t,-1+2t)$.  Then the vector $\vec {v(t)}=P(t)-P=(t,-2t,-3+2t)$ should be perpendicular to $\vec n$ so we want to solve:  $$t-6t-3+2t=0\implies 3t=-3\implies t=-1$$
Thus $M_1$ is given by $$P+t\vec {v(-1)}=(1,2,2)+t(-1,2,-5)$$
Check:  Since $P$ is not in the given plane we should see that no point in $M_1$ is in the plane.  We try to solve $$1-t+6+6t+2-5t=1$$ and confirm that there is no solution.
A: $M_1$ does not have to be on then plane $x + 3y + z = 1$.  If the line $M_1$ is parallel to the plane $x + 3y + z = 1$, then it must be on the plane
$x + 3y + z = k$ for some real number $k$. Since the line $M_1$ contains the point $P=(1,2,2)$, and since $(1) + 3(2) + (2) = 9$,  then the line is on the plane $x + 3y + z = 9$.
The plane $x + 3y + z = 9$ intersects the line $M_2(t) =  (1+t, 2-2t, -1+2t)$ when
\begin{align}
   (1+t) + 3(2-2t) + (-1+2t) &= 9 \\
   6 - 3t &= 9 \\
   t &= -1
\end{align}
Hence the point $M_2(-1) = (0, 4, -3)$ is also on the line $M_1$. So the line $M_1$ is parallel to the direction 
$(1,2,2) - (0, 4, -3) = (1,-2,5)$, then the equation of $M_1$ is
$$M_1(t) = (1,2,2) + t(1,-2,5)$$
We can verify your three requirements


*
  
*$M_1$ goes through the point $P(1,2,2).$
    
*
      
*Because $M_1(0) = (1,2,2)$.
    
 
  
*Is parallel with the plane $x + 3y + z = 1.$
    
*
      
*Because $x+3y+z=(1+t)+3(2-2t)+(2+5t) = 9.$
    
  
  
*Has an intersection somewhere with the line $M_2 = 1+t, 2-2t, -1+2t.$
    
*
      
*Because $M_2(-1) = (0,4,-3)=M_1(-1)$
