How to determine $x, y, z$ coordinates of the third vertex of a 3D triangle? Given a line $AB$ such that $A(0,0,0)$ and $B(4,7,9)$. How can I obtain a point $C(x,y,z)$ of $\Delta ABC$ with $AB$, $AC$, and $BC$ known?
Any help will be appreciated!

We are given two fixed points  $A$ and $B$ in three space and two positive numbers $a$ and $b$ which together with $c=|AB|$ satisfy the triangle inequalities. We are looking for a point $C$ such that $|AB| = c, \,\, |BC|=a$ and $|CA| = b$. 
Assume we have found one such point $C$. Let $H \in AB$ be the (unique) point on $AB$ such that $CH$ is perpendicular to $AB$. Since we are in three space, it turns out that all points $C$ that satisfy the conditions $|BC|=a$ and $|CA| = b$ form a circle lying on the plane through $H$ perpendicular to line $AB$. Moreover, the center of the circle is $H$ and its radius is $|CH|$. WE want to find a parametrization of that circle.
As already mentioned, point $H \in AB$ is such that $CH$ is perpendicular to $AB$, i.e. $CH$ is the altitude of triangle $ABC$ through the vertex $C$. Then $H$ splits the segment $AB$ into two segments $AH$ and $BH$. Let $|AH| = c_b$ and $|BH| = c_a$. By construction $c_a+c_b = c.$ Furthermore, by Pythagoras' theorem $$b^2 - c_b^2 = |CH|^2 = a^2-c_a^2.$$ Thus, we have a system of 2 equations for the unknown variables $c_a,c_b$:
\begin{align}
c_b^2-c_a^2 &= b^2-a^2\\
c_b+c_a &= c
\end{align} 
and when we solve it we obtain 
\begin{align}
c_b &= \frac{b^2-a^2+c^2}{2c}\\
c_a &= \frac{a^2-b^2+c^2}{2c}
\end{align}
Furthermore, we can find the length of the altitude
$$|CH| = \sqrt{b^2-c_b^2} = \frac{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}{2c}$$
Next, observe that the coordinate system is with origin $A$. Let $\overrightarrow{e}_1 = (1,0,0)$. Then clearly $\overrightarrow{e}_1$ and $\overrightarrow{AB}$ are linearly independent. Therefore vector $\overrightarrow{e}_1 \times \overrightarrow{AB}$ is perpendicular to vector $\overrightarrow{AB}$ and vector $\overrightarrow{AB} \times(\overrightarrow{AB} \times \overrightarrow{e}_1)$ is perpendicular to both vectors $\overrightarrow{AB}$  and $\overrightarrow{e}_1 \times \overrightarrow{AB}$. Furthermore, the vectors $$\frac{\overrightarrow{e}_1 \times \overrightarrow{AB}}{|\overrightarrow{e}_1 \times \overrightarrow{AB}|} \,\,\, \text{ and }\,\,\, \frac{\overrightarrow{AB} \times(\overrightarrow{AB} \times \overrightarrow{e}_1)}{|\overrightarrow{AB} \times(\overrightarrow{AB} \times \overrightarrow{e}_1)|}$$ are of unit length and are perpendicular to each other, as well as each of them is perpendicular to $\overrightarrow{AB}$. Therefore, the circle we are looking for can be parametrized as
$$\overrightarrow{AC} = \overrightarrow{AH} + |CH| \left(\cos{\theta} \, \frac{\overrightarrow{AB} \times(\overrightarrow{AB} \times \overrightarrow{e}_1)}{|\overrightarrow{AB} \times(\overrightarrow{AB} \times \overrightarrow{e}_1)|} + \sin{\theta} \, \frac{\overrightarrow{e}_1 \times \overrightarrow{AB}}{|\overrightarrow{e}_1 \times \overrightarrow{AB}|}\right).$$ But $$\overrightarrow{AH} = c_b\frac{\overrightarrow{AB}}{|AB|} = \frac{c_b}{c}\overrightarrow{AB}.$$ Thus
$$\overrightarrow{AC} = \frac{c_b}{c}\overrightarrow{AB} + \sqrt{c^2-c_b^2} \left(\cos{\theta} \, \frac{\overrightarrow{AB} \times(\overrightarrow{AB} \times \overrightarrow{e}_1)}{|\overrightarrow{AB} \times(\overrightarrow{AB} \times \overrightarrow{e}_1)|} + \sin{\theta} \, \frac{\overrightarrow{e}_1 \times \overrightarrow{AB}}{|\overrightarrow{e}_1 \times \overrightarrow{AB}|}\right).$$ We know everything in this equation -- we have vectors $\overrightarrow{AB}$ and $\overrightarrow{e_1}$, as well as 
\begin{align}
c_b &= \frac{b^2-a^2+c^2}{2c}\\
\sqrt{b^2-c_b^2} &= \frac{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}{2c}.
\end{align}
My brief computations show that in numbers
$$
\begin{pmatrix}
C_1\\ C_2 \\ C_3
\end{pmatrix} =  \frac{c_b}{c}\begin{pmatrix}
4\\  7\\ 9 
\end{pmatrix}  + \sqrt{c^2-c_b^2} \left(\frac{\cos{\theta}}{\sqrt{4749}}  \, \begin{pmatrix}
-65\\  14\\ 18 
\end{pmatrix}  + \frac{\sin{\theta}}{\sqrt{130}}  \,  \begin{pmatrix}
0\\  -9\\ 7 
\end{pmatrix}\right) 
$$
