Answer
$x-2y-2z=3$
Work Step by Step
The normal vector tangent to the plane is:
$n=- i+2v j+ 2u k$
The point $( -1,-1,-1)$ corresponds to the parameter values as: $r( -1,-1,-1) =-i+2j+2k$
When $a$ is the position vector of a point on the plane and $n$ is the vector normal to the plane, then the equation of the plane is: $(r-a) \cdot n=0$
Therefore, $(x +1) \cdot (-1)+(y+1) \cdot (2) +(z+1) \cdot (2)=0$
$\implies -x-1+2y+2+2z+2=0$
$\implies x-2y-2z=3$
