Can I calculate this coordinate with the given?

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I studying about the total station.
This time, I want to know how the total station calculate the coordinate of its point with two known points.

Their coordinates in the global coordinate system are $$X (\alpha, \beta, \gamma)$$$$P_1 (a_1, b_1, c_1)$$$$P_2 (d_1, e_1, f_1)$$ and in the local coordinate system origin at X are $$ X (0, 0, 0) $$ $$ P_1 (l_1\sin\theta_1\cos\phi_1, l_1\sin\theta_1\sin\phi_1, l_1\cos\theta_1) $$ $$ P_2 (l_2\sin\theta_2\cos\phi_2, l_2\sin\theta_2\sin\phi_2, l_2\cos\theta_2) $$ Except the coordinate of $X$, others are all already known.

How can I calculate the coordinate of $X$ with these?

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The coordinates in the global (world) coordinate system is given by

$ p = X + R q $

where $p$ is the global coordinate vector and $q$ is the local coordinate vector.

We know that

$P_1 = X + R Q_1 $

$P_2 = X + R Q_2 $

Subtract,

$P_2 - P_1 = R (Q_2 - Q_1) \hspace{30pt} (1)$

This is only one equation involving $R$ without $X$, so $R$ cannot be determined, we need one extra point

$P_3 = X + R Q_3 $

Then we can compute the difference,

$P_3 - P_1 = R (Q_3 - Q_1) \hspace{30pt} (2)$

Using $(1)$ and $(2)$ we can determine a unique $R$ as follows

$(P_3 - P_1) \times (P_2 - P_1) = ( R (Q_3 - Q_1) ) \times ( R(Q_2 - Q_1) ) = R ( (Q_3 - Q_1) \times (Q_2 - Q_1) ) \hspace{30pt} (3) $

Combining $(1), (2), (3) $ we get the matrix equation

$ A = R B $

where

$A = [ (P_3 - P_1) , (P_2 - P_1), (P_3 - P_1) \times (P_2 - P_1) ] $

$B = [ (Q_3 - Q_1), (Q_2 - Q_1), (Q_3 - Q_1) \times (Q_2 - Q_1 ) ] $

Then

$ R = A B^{-1} $

Substitute this $R$ matrix in

$ P_1 = X + R Q_1 $

and solve for $X$.