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I'm given $6$ side lengths and asked to find the volume. I feel lost

Tetrahedron $ABCD$ has $AB=1.5$, $AC=1$, $BC=2$, $BD=2.5$, $AD=1.5$, and $CD=3$.

I got the area of triangle ABC by Heron's formula to be $\sqrt{2.25(2.25-1)(2.25-2)(2.25-1.5)}$.

How do I find the height of the tetrahedron?

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    You could use this approach: https://en.wikipedia.org/wiki/Heron's_formula#Heron-type_formula_for_the_volume_of_a_tetrahedron – 2017-02-12
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    Or just notice there is no such tetrahedron, by focusing on $ACD$. – 2017-02-12

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By the Cayley-Menger determinant

$$ 288 V^2=\det\begin{pmatrix}0 & 1 & 1 & 1 & 1 \\ 1 & 0 & d_{12}^2 & d_{13}^3 & d_{14}^2\\ 1 & d_{21}^2 & 0 & d_{23}^2 & d_{24}^2\\1 & d_{31}^2 & d_{32}^2 & 0 & d_{34}^2 \\1 & d_{41}^2 & d_{42}^2 & d_{43}^2 & 0\end{pmatrix} $$ and in our case $d_{12}=d_{21}=\frac{3}{2}$, $d_{13}=d_{31}=1$, $d_{14}=d_{41}=\frac{3}{2}$, $d_{23}=d_{32}=2$, $d_{24}=d_{42}=\frac{5}{2}$, $d_{34}=d_{43}=3$. Since the involved matrix is symmetric but not positive definite, that tells us Houston has a problem: the sides of $ACD$ do not fulfill the triangular inequality, since $AC+AD\color{red}{<}CD$, hence there is no tetrahedron with such side lengths.