How do you find the equation of the plane in xyz-space through the point p=(4 5 4) and perpendicular to the vector n=(-5 -3 -4)?

The answer is: ##5x+3y+4z-51=0##
Given a poiint ##P(x_py_pz_p)## and a vector ##vecv(abc)## perpendicular to the plane the equation is:
##a(x-x_p)+b(y-y_p)+c(z-z_p)=0##
So:
##-5(x-4)-3(y-5)-4(z-4)=0rArr##
##-5x+20-3y+15-4z+16=0rArr##
##5x+3y+4z-51=0##