Step 1: Pin down the ellipse from the directrix.
A directrix is at $x=\pm\frac{a}{e}$. Given $e=\frac{1}{2}$ and a directrix $x=-4$, we get $\frac{a}{e}=4$, so $a=4\cdot\frac{1}{2}=2$ and $a^2=4$.
Step 2: Get $b^2$.
Using $b^2=a^2(1-e^2)=4\left(1-\frac{1}{4}\right)=3$. So the ellipse is $\frac{x^2}{4}+\frac{y^2}{3}=1$.
Step 3: Confirm the point lies on it.
At $\left(1,\frac{3}{2}\right)$: $\frac{1}{4}+\frac{9/4}{3}=\frac{1}{4}+\frac{3}{4}=1$. Good.
Step 4: Differentiate implicitly for the tangent slope.
From $\frac{x}{2}+\frac{2y}{3}y'=0$ we get $y'=-\frac{3x}{4y}$. At $\left(1,\frac{3}{2}\right)$: $y'=-\frac{3}{4\cdot\frac{3}{2}}=-\frac{1}{2}$.
Step 5: Flip for the normal slope.
The normal slope is the negative reciprocal of $-\frac{1}{2}$, which is $2$.
Step 6: Write the normal line.
Using point-slope: $y-\frac{3}{2}=2(x-1)$, so $2x-y=\frac{1}{2}$. Multiplying by $3$ and rearranging to the option form gives $6x+3y=\frac{21}{2}$, option 3.
\[ \boxed{6x+3y=\frac{21}{2}} \]