Where is the mistake?:
SRT:
$$\Delta x'=\Delta x/\gamma$$
$$\Delta y'=\Delta y$$
$$\Delta z'=\Delta z$$
$$\Delta t'=\Delta t\gamma$$
Write down the correct answer.
Look at the separation between two events, A and B, on two distinct world lines with the same velocity. The separation at the same time in one coordinate system, $$\Delta t = 0$$, is not the same as the separation in the other coordinate system, $$\Delta t' = 0$$.
Assume the Lorentz transform applies since we are exploring special relativity. If you don't see a K in an equation, then I have not used any part of special relativity for that result.
$$K=c^{\tiny -2}$$ (eqn 1)
So the coordinate separation between events in one system is:
$${\huge \Delta S} \equiv \begin{pmatrix} \rule{0pt}{20ex} \Delta t \\ \rule{0pt}{20ex} \Delta x \\ \rule{0pt}{20ex} \Delta y \\ \rule{0pt}{20ex} \Delta z \end{pmatrix} \equiv \begin{pmatrix} \rule{0pt}{20ex} t_A - t_B \\ \rule{0pt}{20ex} x_A - x_B\\ \rule{0pt}{20ex} y_A - y_B\\ \rule{0pt}{20ex} z_A - z_B \end{pmatrix} $$ (eqn 2)
The following equation contrains events A and B to be on parallel world lines both moving at velocity $$(u_x, \; u_y, \; u_z)$$ :
$$\begin{pmatrix} \rule{0pt}{20ex} 0 & 0 & 0 & 0 \\ \rule{0pt}{20ex} -u_x & 1 & 0 & 0 \\ \rule{0pt}{20ex} -u_y & 0 & 1 & 0 \\ \rule{0pt}{20ex} -u_z & 0 & 0 & 1 \end{pmatrix} {\huge \Delta S} = \begin{pmatrix} \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} L_x \\ \rule{0pt}{20ex} L_y \\ \rule{0pt}{20ex} L_z \end{pmatrix} $$ (eqn 3)
So the spatial separation between events A and B is ONLY equal to $$(L_x, \; L_y, \; L_z)$$ if $$(u_x = 0, \; u_y = 0, \; u_z = 0)$$ or if $$\Delta t = 0$$:
$$u_x = 0 \, \wedge \, u_y = 0 \, \wedge \, u_z = 0 \; \vee \; \Delta t = 0 \quad \Leftrightarrow \quad \begin{pmatrix} \rule{0pt}{20ex} \Delta x \\ \rule{0pt}{20ex} \Delta y \\ \rule{0pt}{20ex} \Delta z \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} L_x \\ \rule{0pt}{20ex} L_y \\ \rule{0pt}{20ex} L_z \end{pmatrix}$$ (eqn 4)
So now we introduce the Lorentz transform which moves us to a new coordinate system.
$${\huge \Delta S'} \equiv \begin{pmatrix} \rule{0pt}{20ex} \Delta t' \\ \rule{0pt}{20ex} \Delta x' \\ \rule{0pt}{20ex} \Delta y' \\ \rule{0pt}{20ex} \Delta z' \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} \frac{1}{\sqrt{1 - K v^2}} & \frac{K v}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{v}{\sqrt{1 - K v^2}} & \frac{1}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 1 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 0 & 1 \end{pmatrix} {\huge \Delta S} $$ (eqn 5a)
For $$\left| v \right| < c$$ the Lorentz transform is invertible and its inverse is just a Lorentz transform with a velocity of the opposite sign.
$${\huge \Delta S} = \begin{pmatrix} \rule{0pt}{20ex} \frac{1}{\sqrt{1 - K v^2}} & \frac{- K v}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{-v}{\sqrt{1 - K v^2}} & \frac{1}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 1 & 0 \\ \rule{0pt}{20ex} 0 & 0 & 0 & 1 \end{pmatrix} {\huge \Delta S'} $$ (eqn 5b)
From equations 3 and 5b we get:
$$\begin{pmatrix} \rule{0pt}{20ex} 0 & 0 & 0 & 0 \\ \rule{0pt}{20ex} \frac{-(u_x + v)}{\sqrt{1 - K v^2}} & \frac{1 + K u_x v}{\sqrt{1 - K v^2}} & 0 & 0 \\ \rule{0pt}{20ex} \frac{-u_y}{\sqrt{1 - K v^2}} & \frac{K u_y v}{\sqrt{1 - K v^2}} & 1 & 0 \\ \rule{0pt}{20ex} \frac{-u_z}{\sqrt{1 - K v^2}} & \frac{K u_z v}{\sqrt{1 - K v^2}} & 0 & 1 \end{pmatrix} {\huge \Delta S'} = \begin{pmatrix} \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} L_x \\ \rule{0pt}{20ex} L_y \\ \rule{0pt}{20ex} L_z \end{pmatrix}$$ (eqn 6a)
And multiplying both sides on the left by $$ \tiny \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \frac{\sqrt{1 - K v^2}}{1 + K u_x v} & 0 & 0 \\ 0 & \frac{- K u_y v}{1 + K u_x v} & 1 & 0 \\ 0 & \frac{- K u_z v}{1 + K u_x v} & 0 & 1 \end{pmatrix}$$ we recover a constraint similar in form to equation 3.
$$\begin{pmatrix} \rule{0pt}{20ex} 0 & 0 & 0 & 0 \\ \rule{0pt}{20ex} \frac{-(u_x + v)}{1 + K u_x v} & 1 & 0 & 0 \\ \rule{0pt}{20ex} \frac{-u_y \sqrt{1 - K v^2}}{1 + K u_x v} & 0 & 1 & 0 \\ \rule{0pt}{20ex} \frac{-u_z \sqrt{1 - K v^2}}{1 + K u_x v} & 0 & 0 & 1 \end{pmatrix} {\huge \Delta S'} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \frac{\sqrt{1 - K v^2}}{1 + K u_x v} & 0 & 0 \\ 0 & \frac{- K u_y v}{1 + K u_x v} & 1 & 0 \\ 0 & \frac{- K u_z v}{1 + K u_x v} & 0 & 1 \end{pmatrix} \begin{pmatrix} \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} L_x \\ \rule{0pt}{20ex} L_y \\ \rule{0pt}{20ex} L_z \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} \frac{\sqrt{1 - K v^2}}{1 + K u_x v} L_x \\ \rule{0pt}{20ex} L_y - \frac{K u_y v}{1 + K u_x v} L_x\\ \rule{0pt}{20ex} L_z - \frac{K u_z v}{1 + K u_x v} L_x \end{pmatrix}$$ (eqn 6b)
Or $$u'_x = \frac{u_x + v}{1 + K u_x v}, \; u'_y = \frac{u_y \sqrt{1 - K v^2}}{1 + K u_x v}, \; u'_z = \frac{u_z \sqrt{1 - K v^2}}{1 + K u_x v}, \; L'_x = \frac{\sqrt{1 - K v^2}}{1 + K u_x v} L_x \, \; L'_y = L_y - \frac{K u_y v}{1 + K u_x v} L_x , \; \textrm{and} \; L'_z = L_z - \frac{K u_z v}{1 + K u_x v} L_x$$ and:
$$\begin{pmatrix} \rule{0pt}{20ex} 0 & 0 & 0 & 0 \\ \rule{0pt}{20ex} -u'_x & 1 & 0 & 0 \\ \rule{0pt}{20ex} -u'_y & 0 & 1 & 0 \\ \rule{0pt}{20ex} -u'_z & 0 & 0 & 1 \end{pmatrix} {\huge \Delta S'} = \begin{pmatrix} \rule{0pt}{20ex} 0 \\ \rule{0pt}{20ex} L'_x \\ \rule{0pt}{20ex} L'_y \\ \rule{0pt}{20ex} L'_z \end{pmatrix} $$ (eqn 6c)
Such that
$$u'_x = 0 \, \wedge \, u'_y = 0 \, \wedge \, u'_z = 0 \; \vee \; \Delta t' = 0 \quad \Leftrightarrow \quad \begin{pmatrix} \rule{0pt}{20ex} \Delta x' \\ \rule{0pt}{20ex} \Delta y' \\ \rule{0pt}{20ex} \Delta z' \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} L'_x \\ \rule{0pt}{20ex} L'_y \\ \rule{0pt}{20ex} L'_z \end{pmatrix}$$ (eqn 7)
Which is the content of special relativity with respect to length contraction.
So if $$u_x = 0 \, \wedge \, u_y = 0 \, \wedge \, u_z = 0$$ we have $$u'_x = v, \; u'_y = 0, \; u'_z = 0, \; L'_x = \sqrt{1 - K v^2} L_x \, \; L'_y = L_y , \; \textrm{and} \; L'_z = L_z$$ and so it follows that
$$u_x = 0 \, \wedge \, u_y = 0 \, \wedge \, u_z = 0 \, \wedge \; \begin{pmatrix} \rule{0pt}{20ex} \Delta x \\ \rule{0pt}{20ex} \Delta y \\ \rule{0pt}{20ex} \Delta z \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} L_x \\ \rule{0pt}{20ex} L_y \\ \rule{0pt}{20ex} L_z \end{pmatrix} \; \wedge \, \Delta t' = 0 \Rightarrow \begin{pmatrix} \rule{0pt}{20ex} \Delta x' \\ \rule{0pt}{20ex} \Delta y' \\ \rule{0pt}{20ex} \Delta z' \end{pmatrix} = \begin{pmatrix} \rule{0pt}{20ex} \sqrt{1 - K v^2} L_x \\ \rule{0pt}{20ex} L_y \\ \rule{0pt}{20ex} L_z \end{pmatrix} $$ (eqn 8)
which is a specific conclusion, NOT a general rule.
Special relativity is NOT a rescaling of the coordinate axes -- it is much more akin to a rotation of the space and time axis.
Russian Wikipedia on hyperbolic rotation
