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Solution

Given line: $3x-2y+12=0 \Rightarrow y=\frac{3x+12}{2}$
Parabola: $4y=3x^2 \Rightarrow y=\frac{3x^2}{4}$

Equate:
$\frac{3x^2}{4}=\frac{3x+12}{2}$
$\Rightarrow 3x^2=6x+24$
$\Rightarrow x^2-2x-8=0$
$\Rightarrow (x-4)(x+2)=0$

Points: $A(4,12),\ B(-2,3)$
Vertex $V(0,0)$

Slope $VA=\frac{12}{4}=3,\quad VB=\frac{3}{-2}=-\frac{3}{2}$

$\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right| =\left|\frac{3+\frac{3}{2}}{1- \frac{9}{2}}\right| =\left|\frac{\frac{9}{2}}{-\frac{7}{2}}\right| =\frac{9}{7}$

$\theta=\tan^{-1}\left(\frac{9}{7}\right)$

$\boxed{\tan^{-1}\left(\frac{9}{7}\right)}$