Answer
$\langle 3,-8,11\rangle $
Work Step by Step
$3(x-4)-8(y-1)+11z=0$ 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 $
