Answer
$$25x-y+15z=62$$
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: $n_{I}=\lt 2,5, -3 \gt$ and the the normal vector for the second plane as: $n_{II}=\lt-1,5,2 \gt$.
Thus, $n=n_{I} \times n_{II}=\lt 2,5, -3 \gt \times \lt-1,5,2 \gt=\lt 25, -1, 15\gt$
Plug the values $\lt p,q,r\gt=\lt 25, -1, 15 \gt$ into the equation (1) to obtain:
$$ 25(x-0) +(-1) (y+2) +15(z-4)=0 \\ \implies 25x-y+15z=62$$
