$a=3+i$
$z=2-3i$
$az=(3+i)(2-3i)$
$\quad= 6-9 i+2i-3i^2$
$\quad= 6-7i+3=9-7i$
$3az=3(9-7i)+3=9-7i$
$-az= -(9-7i)=-9+7i$
$arg (az)=\tan^{-1} \bigg(\large\frac{-7}{9}\bigg)$
$arg (3az)=\tan^{-1} \bigg(\large\frac{-21}{27}\bigg)$
$\qquad=\tan^{-1} \bigg(\large\frac{-7}{9}\bigg)$
$arg (-az)=\tan^{-1} \bigg(\large\frac{7}{-9}\bigg)$
$\qquad=\tan^{-1} \bigg(\large\frac{-7}{9}\bigg)$
The complex numbers $az,3az,-az$ are complex numbers with same amplitude.
$az,3az,-az$ lie o a straight line.
Hence 4 is the correct answer.