Hint: \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)

Step 1: Find the position of the final image.
Object distance, u = -14 cm
Focal length of the concave lens, f = -21 cm
Using the lens formula:

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$
$\frac{1}{v}=\frac{1}{21}-\frac{1}{14}=\frac{-2-3}{42}=\frac{-5}{42}$
$v=\frac{-42}{5}=-8.4 cm$
The image is formed 8.4 cm away from the lens on the right side. The negative sign shows that the image is erect and virtual. 

Step 2: Find the magnification of the image.
The magnification of the image:

$m=\frac{\mathrm{Im}age height \left(h_2\right)}{Object height \left(h_1\right)}=\frac{v}{u}$
$\Rightarrow h_2=\frac{-8.4}{-14}\times 3=1.8 cm.$
Hence, the height of the image is 1.8 cm.
If the object moves further away from the lens, then the virtual image will move towards the focus of the lens, but not beyond it. The size of the image will decrease with the increase in the object distance.