Your equations are wrong. The \$V_3\$ branch is not considered since the switch is open. If the switch was open, you would then have 3 loops in total, and \$I_1\$ would be split.
Following your assumptions about the currents:
$$I_2 = I_1 + I_3$$
\$I_1\$ can be gotten right away since you somehow know that the voltage across \$R_7\$ (top to bottom) is \$15 V \$.
$$I_1 = 15/50 = 0.3 A$$
$$I_2 = 0.3 + I_3 ............(1)$$
The equation for the first loop is:
$$-20I_2 - V_2 - 75I_3 = 0$$
$$20I_2 + V_2 = -75I_3..........(2)$$
The equation for the second loop is:
$$80I_1 + V_2 + 20I_2 = 0$$
$$24 + V_2 + 20I_2 = 0$$
$$20I_2 + V_2 = -24............(3)$$
Equating (2) and (3) gives:
$$-24 = -75I_3$$
$$I_3 = 0.32 A$$
$$I_2 = 0.3 + 0.32 = 0.62 A$$
$$V_2 = -24 - 20I_2 = -24 - 12.4 = -36.4 V$$
Meaning, the polarity of \$V_2\$ in your schematic is the other way round.