## Calculus (3rd Edition)

$\langle 3,-8,11\rangle$
$3(x-4)-8(y-1)+11z=0$ can be rewritten as $3x-8y+11z=4$. The normal vector of a plane in the form $ax+by+cz=d$ is $\vec n=\langle a,b,c\rangle$ Therefore, the normal vector of $3x-8y+11z=4$ is given as: $\vec n=\langle 3,-8,11\rangle$