Answer
$-4x+3y+z=-14$
Work Step by Step
Given: $(3,-1,1),(4,0,2),(6,3,1)$
Use the points to make two vectors parallel to the plane.
$a=\lt 4-3,0-(-1),2-1\gt=\lt 1,1,1\gt$
$b=\lt 6-3,3-(-1),1-1\gt=\lt 3,4,0\gt$
Use cross product to make a vector perpendicular to these vectors (and to the plane)
$n=a\times b=\lt-4,3,1\gt$
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 $(4,0,2)$ and has normal vector $=\lt -4,3,1\gt$
Therefore,
$\lt -4,3,1\gt$.$(x-4,y-0,z-2)=0$
A scalar equation is:
$-4(x-4)+3y+1(z-2)=0$
$-4x+16+3y+z-2=0$
Hence, the required equation of the plane is $-4x+3y+z=-14$
