I'm currently solving the exercises of Modern Robotics book. I came across this problem but it is not clear to me how to approach it. I computed the analytic Jacobian which is

$$
\begin{bmatrix} \dot{x}\\\dot{y} \\ \dot{z}\end{bmatrix} =
\left(\begin{array}{ccc} L\,s\_{\theta\_1}\,s\_{\theta\_2}\,s\_{\theta\_3}-L\,c\_{\theta\_2}\,s\_{\theta\_1}-L\,c\_{\theta\_2}\,c\_{\theta\_3}\,s\_{\theta\_1}-L\,c\_{\theta\_1} & -L\,c\_{\theta\_1}\,\left(s\_{\theta\_2}+\sin\left(\theta\_2+\theta\_3\right)\right) & -L\,c\_{\theta\_1}\,\sin\left(\theta\_2+\theta\_3\right)\\ L\,c\_{\theta\_1}\,c\_{\theta\_2}-L\,s\_{\theta\_1}+L\,c\_{\theta\_1}\,c\_{\theta\_2}\,c\_{\theta\_3}-L\,c\_{\theta\_1}\,s\_{\theta\_2}\,s\_{\theta\_3} & -L\,s\_{\theta\_1}\,\left(s\_{\theta\_2}+\sin\left(\theta\_2+\theta\_3\right)\right) & -L\,s\_{\theta\_1}\,\sin\left(\theta\_2+\theta\_3\right)\\ 0 & L\,\left(c\_{\theta\_2}+\cos\left(\theta\_2+\theta\_3\right)\right) & L\,\cos\left(\theta\_2+\theta\_3\right) \end{array}\right)
\begin{bmatrix} \dot{\theta}\_1\\\dot{\theta}\_2 \\ \dot{\theta}\_3\end{bmatrix}
$$
At zero configuration $\theta\_1=\theta\_2=\theta\_3=0.$, we get
$$
\begin{bmatrix} \dot{x}\\\dot{y} \\ \dot{z}\end{bmatrix} =
\begin{bmatrix} -L&0&0\\2L&0&0\\0&2L&L\end{bmatrix}
\begin{bmatrix} \dot{\theta}\_1\\\dot{\theta}\_2 \\ \dot{\theta}\_3\end{bmatrix}
$$
The analytic Jacobian is not invertible at this configuration. My conclusion is the motion is not possible but it seems to me according to the question, there is a possibility. Any suggestions?

[![enter image description here](https://i.stack.imgur.com/lVVul.png)](https://i.stack.imgur.com/lVVul.png)

[![enter image description here](https://i.stack.imgur.com/z8Hrq.png)](https://i.stack.imgur.com/z8Hrq.png)