Answer
$7x-3y-5z=-14$
Work Step by Step
The normal to the plane is $n=\lt 7,-3,-5 \gt$
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 the point $(1,2,3 )$, we have
$7(x-1)-3(y-2)-5(z-3)=0$
or, $7x-7-3y+6-5z+15=0$
Hence, $7x-3y-5z=-14$
