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Fiddling

 * $$\begin{align}

-2x &&\; + \;&& y &&\; +\;&& 2z &&\; = \;&& -3 &  \qquad (L_3) \text {xx} \end{align}$$

Calculation



 * $$ \ell = \ell_1 + \ell_2, \tan \alpha = \frac{d}{\ell_1} ~ \Rightarrow ~ \ell_1 = \frac{d}{\tan \alpha}$$


 * $$ \ell = \frac{d}{\tan \alpha} + \frac{d}{\tan \beta}$$

Therefore


 * $$d = \ell \, / \, (\tfrac{1}{\tan \alpha} + \tfrac{1}{\tan \beta})$$

Using the trigonometric identities tan α = sin α / cos α and sin(α + β) = sin α cos β + cos α sin β, this is equivalent to:


 * $$d = \frac{\ell \sin\alpha \sin\beta}{\sin(\alpha + \beta)}$$

From this, it is easy to determine the distance of the unknown point from either observation point, its north/south and east/west offsets from the observation point, and finally its full coordinates.