1. Write both conics in standard form

The hyperbola is $H:\frac{x^2}{a^2}-y^2=1$ which can be written as $\frac{x^2}{a^2}-\frac{y^2}{1}=1.$ So for the hyperbola, $b_H^2=1.$

The ellipse is $E:3x^2+4y^2=12.$ Dividing by $12$, $\frac{x^2}{4}+\frac{y^2}{3}=1.$ So for the ellipse, $a_E^2=4,\quad b_E^2=3.$

2. Use latus rectum lengths

For a hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, the length of latus rectum is $L_H=\frac{2b^2}{a}.$ Here $b_H^2=1$, so $L_H=\frac{2}{a}.$

For an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, the length of latus rectum is $L_E=\frac{2b^2}{a}.$ Here $a_E=2$ and $b_E^2=3$, so $L_E=\frac{2\cdot 3}{2}=3.$

Given $L_H=L_E$, $\frac{2}{a}=3 \implies a=\frac{2}{3}.$

Thus for the hyperbola, $a_H^2=\frac{4}{9},\quad b_H^2=1.$

3. Find eccentricity of the hyperbola

For the hyperbola, $e_H^2=1+\frac{b_H^2}{a_H^2}.$ So, $e_H^2=1+\frac{1}{4/9}=1+\frac{9}{4}=\frac{13}{4}.$

4. Find eccentricity of the ellipse

For the ellipse, $e_E^2=1-\frac{b_E^2}{a_E^2}=1-\frac{3}{4}=\frac{1}{4}.$

5. Compute the required value

$e_H^2+e_E^2=\frac{13}{4}+\frac{1}{4}=\frac{14}{4}=\frac{7}{2}.$

Therefore, $12\left(e_H^2+e_E^2\right)=12\cdot \frac{7}{2}=42.$

6. Comparison with stored answer

Derived answer = $42$.

Stored correct answer = $42$.

They agree.