from below.

$|V_o|$ will increase to $2(\frac{15}{12})=2\mathrm{.}5V.$

$D_1$ is reverse$-$biased, so increasing $V_{CC}$ will decrease $|VOI|$ to $2(\frac{12}{15})=1\mathrm{.}6V.$

Since $I_D=I_S{e}^{(\frac{V_D}{V_T})}$ for a diode, increasing the voltage by a factor $\frac{15}{12}=1\mathrm{.}25,$ means $|V_0|$ will increase by a factor $e^{1\mathrm{.}25}=3\mathrm{.}49.$

IVol will stay essentially the same, namely, $2$ V $.$