Answer
$(1,3)$, see graph.
Work Step by Step
Step 1. Start from the original and form squares with variables: $3x^2-6x+4y^2-24y=-39$ or $3(x^2-2x+1)+4(y^2-6y+9)=3+36-39$ which gives $3(x-1)^2+4(y-3)^2=0$
Step 2. As both squares are larger or equal to zero, the above equation represents a degenerate conic of a point $(1,3)$
Step 3. See graph.
