Answer
$-6x-3y+3z=-3$
Work Step by Step
Since, we have $t=2s+2$ and $-t+2=s+3$
After solving, we get $s=-1$ and $t=2s
+2=2(-1)+2=0$
The normal to the plane is $n=\lt -6,-3,3 \gt$
We know that the standard equation of a plane passing through the point $(x_0,y_0,z_0)$ is written as: $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$
Then for point $(0,2,1)$, we have
$-6(x-0)-3(y -2)+3(z-1)=0$
or, $-6x-3y+3z=-3$
