Given that the triangle is dilated by factor 3, the image will be found as follows;
The object is at:
A(-7,-3), B(-3,-2), C(-4,-5)
when the image was enlarged the new coordinates will be:
A'=3(-7,-3)=(-21,-9)
B'=3(-3,-2)=(-9,-6)
C'=3(-4,-5)=(-12,-15)
since the image is centered at the point (-7,-6), the final point will be at:
A"=[(-21+-7),(-9+-6)=(-28,-15)
B''=[(-9+-7),(-6+-6)]=(-16,-12)
C''=[(-12+-7),(-15+-6)]=(-19,-17)
thus the coordinates of the final image are:
A''(-28,-15),B''(-16,-12),C''(-19,-17)