Answer
there is no non trivial way to write $0$ as a linear combination of the given vectors.
Work Step by Step
Suppose the following linear combination
$$a(1,0,1)+b(-1,1,2)+c(0,1,4)=(0,0,0), \quad a,b,c\in R.$$
Which yields the following system of equations
\begin{align*}
a-b&=0\\
b+c&=0\\
a+2b+4c&=0.
\end{align*}
The determinent coefficient matrix is given by $$
\left|\begin{array}{rrrr}{1} & {-1} & {0} \\ {0} & {1} & {1} \\ {1} & {2} & {4} \end{array}\right|=\left|\begin{array}{rrrr}{1} & {-1} & {0} \\ {0} & {1} & {1} \\ {0} & {3} & {4} \end{array}\right|=1.
$$
Since the determinent of the coefficient matrix is non zero then the system has only the trivial solution, that is, $a=0$, $b=0$, $c=0$. Hence, there is no non trivial way to write $0$ as a linear combination of the given vectors.
