#### Answer

$6x+9y-z=26$

#### Work Step by Step

Given: The plane through $(1,2,-2)$ that contains the line $x=2t,y=3-t,z=1+3t$
The direction vector to the line is $a=\lt2,-1,3\gt $
At $t=0$ , the points on the line are $(0,3,1)$.
Use these points to make vectors originating at the given point.
$b=\lt0-1,3-2,1-(-2)\gt=\lt -1,1,-3\gt$
Use cross product to make a vector perpendicular to these vectors (and to the plane)
$n=a\times b=\lt-6,-9,1\gt$
Let $(x_{0},y_{0},z_{0})$ be a point on the plane and $\lt a,b,c\gt$ be a normal vector to the plane.
Then a vector equation of the plane is $\lt a,b,c\gt$.$(x_{0},y_{0},z_{0})=0$
A scalar equation is: $a(x-x_{0}),b(y-y_{0}),c(z-z_{0})$
Since, the plane $(0,3,1)$ and has normal vector $\lt -6, -9, 1\gt$
Therefore,
$\lt -6, -9, 1\gt$.$(x-0,y-3,z-1)=0$
A scalar equation is:
$-6(x-0)-9(y-3)+(z-1)=0$
$-6x-9y+27+z-1=0$
$-6x-9y+z=-26$
Hence, the required equation of the plane is $6x+9y-z=26$