Answer
$7x-3y-5z=-14$
Work Step by Step
The standard equation of a plane passing through the point $(x_1,y_1,z_1)$ can be defined as:
The equation of a normal to the plane is given as: $n=\lt 7,-3,-5 \gt$
Now, the parametric equations for point $(1,2,3 )$ are:
$7(x-1)-3(y-2)-5(z-3)=0$
or, $7x-7-3y+6-5z+15=0 \implies 7x-3y-5z=-14$
