Answer
$$ 6x-4y+z=d$$
Work Step by Step
Write the general equation of a plane in $R^3$ passing through the point $P=(x_0,y_0, z_0)$ along with the normal vector $n=\lt p,q,r\gt$ such that:
$$p(x - x_0) + q(y - y_0) + r(z - z_0) = 0~~~~(1)$$
In other words, the equation of a plane can be described as: $d=px+qy+rz ~~~(2)$
Here, we have the normal vector for the first plane as: $n_{I}=\lt 2,3, 0 \gt$ and the the normal vector for the second plane as: $n_{II}=\lt-1,-1, 2 \gt$.
Thus, $n=n_{I} \times n_{II}=\lt 2,3, 0 \gt \times \lt-1,-1, 2 \gt=\lt 6, -4, 1\gt$
Plug the values $\lt p,q,r\gt=\lt 6, -4, 1\gt$ into the equation (2) to obtain:
$$ 6x-4y+z=d$$
