The relative speed of two objects is the speed of one object in the rest frame of the other object. It's best to work with manifestly covariant objects. In this case these are the proper four-velocities, which are in your case
$$u_A=\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \quad u_B=\frac{1}{\sqrt{1-0.7^2}} \begin{pmatrix}1 \\ -0.7 \\ 0 \\ 0 \end{pmatrix}, \quad u_C=\frac{1}{\sqrt{1-0.7^2}} \begin{pmatrix}1 \\ 0.7 \\ 0 \\ 0 \end{pmatrix}.$$
Now you don't need a Lorentz transformation to get the relative speed of each observer since you can get this with Minkowski products between these four-velocities.
For observer A, at rest in the computational frame, it's easy to see that you get the time-component of the four-velocities of B and C just by the Minkowski product of their four-velocities with ##u_A##:
$$(\gamma_B)_A =u_A \cdot u_B=\frac{1}{\sqrt{1-0.7^2}}.$$
This is the ##\gamma## factor of B measured by A who is at rest in the computational frame. From the ##\gamma## factor you get back ##\beta=|\vec{\beta}|=|\vec{v}/c|## simply by
$$(\beta_B)_A=\sqrt{1-\frac{1}{(\gamma_B)_A^2}}=0.7.$$
That's trivial, but the magic of covariant treatments is that since it's working with invariants, it's a general valid formula, i.e., to get the speed of ##B## as measured by ##C## you simply calculate the ##\gamma## factor of ##B## as measured by ##C## via the Minkowski product of the four-velocities,
$$(\gamma_B)_C=u_C \cdot u_B=\frac{1}{1-0.7^2}(1+0.7^2)=149/51$$
and thus the relative speed
$$(\beta_B)_C=\sqrt{1-1/(\gamma_B)_C^2} \simeq 0.940.$$
