Background
Let's say we have a tetrahedron spanned by the points $A(0, 0, 0), B(5, 0, 0), C(4, 2, 0), T(0, 0, 5)$, and we want to find the volume.
Scenario 1
Ok, one way to do it is to use $\triangle ABC$ as base, and then $AT$ becomes the height of the triangle, since A is the closes point to T on the base. We can then use the formula for the volume of a pyramid, and we get $$V = \frac13 Bh = \frac13\left( \overbrace{\frac12|\vec{AB}\times\vec{AC}|}^{\text{area of triangle base}}\right) \cdot \overbrace{|\vec{AT}|}^{\text{height of figure}} = \frac{25}3$$
Now here, I got "lucky" because $\vec{AT}$ happens to already be the height of the tetrahedron. I didn't have to find the height by connecting the top to the closest point in the triangle base.
Scenario 2
Let's use $\triangle BCT$ as the base, and $A$ as the top.
By the formula for the volume of a tetrahedron with vectors, we get
$$V = \frac16 \left| \overbrace{ \left(\vec{BC} \times \vec{BT}\right)}^{\text{area of triangle base, but 1/2 is accounted for in 1/6}} \cdot \overbrace{\vec{BA}}^{\color{red}{\text{not height of figure}}}\right| = \ldots = \frac{25}{3}$$
Question
In the last scenario, I see that the $\frac16$ arrives by $\frac13$ from the volume of a pyramid, times $\frac12$ from the area of the triangle base. But how come we could just use $\vec{BA}$ like that? It's not the height of the tetrahedron, if we consider $\triangle BCT$ as the base.
From what I can tell, $|\vec{BA}| = 5$ while the distance from $A$ to $\triangle BCT$ is $\frac{10}3$.
