Answer:
Distance: -30.0 cm; image is virtual, upright, enlarged
Explanation:
We can find the distance of the image using the lens equation:

where:
f = 15.0 cm is the focal length of the lens (positive for a converging lens)
p = 10.0 cm is the distance of the object from the lens
q is the distance of the image from the lens
Solving for q,

The negative sign tells us that the image is virtual (on the same side of the object, and it cannot be projected on a screen).
The magnification can be found as

The magnification gives us the ratio of the size of the image to that of the object: since here |M| = 3, this means that the image is 3 times larger than the object.
Also, the fact that the magnification is positive tells us that the image is upright.