The question is missing, but I guess the problem is asking for the distance and the size of the image.
Let's find the distance first, by using the lens equation:
[tex] \frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i} [/tex]
where f is the focal length, [tex]d_o[/tex] the distance of the object from the lens, [tex]d_i[/tex] the distance of the image from the lens. Using [tex]f=-12.0 cm[/tex] and [tex]d_o=3.0 cm[/tex], we find
[tex]- \frac{1}{12}= \frac{1}{3}+ \frac{1}{d_i} [/tex]
[tex]d_i=- \frac{12}{5}=-2.4 cm [/tex]
where the negative sign means the image is virtual, so located in front of the lens.
Now we can find the size of the image, using the relationship:
[tex] \frac{h_i}{h_o} =- \frac{d_i}{d_o} [/tex]
where [tex] h_i[/tex] and [tex] h_o[/tex] are the size of the image and of the object, respectively.
By using [tex] h_o=9.0 cm[/tex], we can find [tex] h_i[/tex], the size of the image:
[tex] h_i=-h_o \frac{d_i}{d_o}=-(9.0 cm) \frac{(-2.4 cm)}{3.0 cm}=7.2 cm [/tex]
where the positive sign means the image is upright.
