According to the question;
Object distance = -12cm;
Image distance = v;
Focal length = 18cm;
By lens formula;
\(\frac1v-\frac1u=\frac1f\)
⇒ \(\frac1v-\frac1{-12}=\frac1{18}\)
⇒ \(\frac1v=\frac1{18}-\frac1{12}\)
⇒ \(\frac1v=\frac{2-3}{36}-\frac{-1}{36}\)
⇒ \(\frac1v=\frac{-1}{36}\)
⇒ v = -36cm.
Therefore, image is formed at 36 cm in front of lens.
Now;
Height of object h1= 5cm;
Magnification = \(\frac{h_2}{h_1}=\frac{v}u\)
Putting values of v and u
Magnification = \(\frac{h_2}{5}=\frac{-36}{-12}\)
⇒ \(\frac{h_2}{5}=3;\)
⇒ h2 = 3 x 5 = 15
Height of image is 15 cm.
Thus, the image is virtual, enlarged, and erect and three times the size of object.