Equation of a plane through line and point Write the equation of the plane that passes through the line of the intersection of the planes $P_1:x-2y+3z=0$ and $P_2: 2x+z-3=0$ and through the point M(-1,2,6)
I take the two planes and I try to find the parametric equations of the line they intersect in. I get: $$x=\frac{-2t-9}{5}\:;y=t\:;z=\frac{4t-3}{5}$$
Then I substitute t = 0 and t =1 to get two more points that are on the line.
I get $A\left(-\frac{9}{5},\:0,\:-\frac{3}{5}\right);B\left(-\frac{11}{5},1,\frac{1}{5}\right)$.
Then I form the vectors MA and MB and I do MA x MB to get the normal vector.
The result I'm getting is $5i+\frac{82}{25}j-\frac{8}{5}k$.
Then I input the point M into the normal vector equation and I get $125x+82y-40z+201 = 0$ as the plane equation, which is definitely wrong, since my answer key says it's a different result.
What am I doing wrong? I'm new at this...

You have an error of sign in the line equation that is:
$$
(x,y,z)=\left(\frac{9-2t}{5},t,\frac{4t-3}{5} \right)
$$
Some suggestion: 
1)your chose for $t$ is not the best. It's more simple $z=t$.
2) You can find the director vector for the common line as the cross product of the normal vector of the two planes: $(1,-2,3)^T \times (2,0,1)^T$ and use a common point of the two planes to find the equation of the common line.
3) To find the final plane it is not necessary to know the equation of the common line. You can find two common point of the two planes in a simpler way. Chose some value for one of the variables, say $x=0$, from the second plane you have $z=3$ and from the first you find $y=9/2$, so $A=(0,9/2,3)$ is a common point. Do the same for $z=-1$ and you find $x=2$ and $y=-1/2$ so another common point is $B=(2,-1/2,-1)$ . Now, with the given point $M$ you have three points and you can find the equation of the plane.
