1. Write the ellipse in standard form

Given ellipse: $3x^2+4y^2=12$ Divide by $12$: $\frac{x^2}{4}+\frac{y^2}{3}=1$ So, $a^2=4,\quad b^2=3$ Since $a^2>b^2$, the major axis is along the $x$-axis.

Its foci are at: $c^2=a^2-b^2=4-3=1 \implies c=1$ Hence the foci are: $(\pm 1,0)$

2. Form the hyperbola with same foci

The hyperbola has the same foci $(\pm 1,0)$, so it must be of the form: $\frac{x^2}{A^2}-\frac{y^2}{B^2}=1$ with $c^2=A^2+B^2=1$

Given transverse axis length $=\sqrt 2$. For hyperbola, transverse axis length $=2A$. So, $2A=\sqrt 2 \implies A=\frac{1}{\sqrt 2}$ Therefore, $A^2=\frac12$ Now, $c^2=A^2+B^2$ $1=\frac12+B^2$ $B^2=\frac12$

Thus the hyperbola is: $\frac{x^2}{1/2}-\frac{y^2}{1/2}=1$ which simplifies to $2x^2-2y^2=1$ or $x^2-y^2=\frac12$

3. Check each option

We test whether each point satisfies $x^2-y^2=\frac12$

Option A: $\left(1,-\frac{1}{\sqrt2}\right)$

$x^2-y^2=1-\frac12=\frac12$ So A lies on the hyperbola.

Option B: $\left(\sqrt{\frac32},\frac{1}{\sqrt2}\right)$

$x^2-y^2=\frac32-\frac12=1$ Since $1\ne \frac12$, this point does not lie on the hyperbola.

Option C: $\left(-\sqrt{\frac32},1\right)$

$x^2-y^2=\frac32-1=\frac12$ So C lies on the hyperbola.

Option D: $\left(\frac{1}{\sqrt2},0\right)$

$x^2-y^2=\frac12-0=\frac12$ So D lies on the hyperbola.

4. Conclusion

The hyperbola does not pass through: $\boxed{\left(\sqrt{\frac32},\frac{1}{\sqrt2}\right)}$ So the correct option is B.