Let $p(z)=a z^{3}+4 z^{2}+3 z-4$

And $\quad q(z)=z^{3}-4 z+a$

As these two polynomials leave the same remainder, when divided by $z-3,$ then $p (3)=$ $q (3).$

$p(3)=a(3)^{3}+4(3)^{2}+3(3)-4$

$=27 a+36+9-4$

Or $\quad p(3)=27 a+41$

And $\quad q(3)=(3)^{3}-4(3)+a$

$=27-12+a=15+a$

Now, $\quad p(3)=q(3)$

$\Rightarrow \quad 27 a+41=15+a$

$\Rightarrow \quad 26 a=-26 a ; a=-1$

Hence, the required value of $a=-1$