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_p,y_p,z_p)## and a vector ##vecv(a,b,c)## 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##