find range for dependent 
There are three vector $a$, $b$, $c$ in three-dimensional real vector space, and the inner product between them $a\cdot a=b\cdot b = a\cdot c= 1, a\cdot b= 0, c\cdot c= 4$. When setting $x = b\cdot c$, answer the following question: when $a, b, c$ are linearly dependent, find all possible values ​of $x$.
  (dot here means dot product)

For dependent condition 
$$\begin{align}
(a×b)·c &= 0\\
a·(b×c) &= 0\\
a(bc \sin θ)&=0
\end{align}$$
So $\theta = 0, \pi$.
Then
$$\begin{align}
x&=|b||c| \cos \theta \\
x&=2 \cos \theta \\
\implies x &= 2 \cos 0, x = 2 \cos \pi \\
x &= \mp 2
\end{align}$$
Am I right using triple cross product? or should i find $\theta$?


Linear dependence places the vectors in the same plane. This makes the problem easier because we can focus on the angle.
From $c\cdot c=4$ we get that $|c|=2$.
So $a\cdot c=2\cos\theta=1\implies\cos\theta=\frac{1}{2}\implies\theta=60^{\circ}$.
From $b\cdot b=1$ and $a\cdot b=0$ we get that $b$ is a unit vector perpendicular to $a$.
Since the angle between $a$ and $c$ is $60^{\circ}$, the angle between $b$ and $c$ is either $30^{\circ}$ or $150^{\circ}$.
Therefore, $x=2\cos{30^{\circ}}$ or $x=2\cos{150^{\circ}}$ so $x=\pm \sqrt{3}$
