Answer
$$(1,0,1)+(-1,1,2)-(0,1,3)=(0,0,0).$$
Work Step by Step
Suppose the following linear combination
$$a(1,0,1)+b(-1,1,2)+c(0,1,3)=(0,0,0), \quad a,b,c\in R.$$
Which yields the following system of equations
\begin{align*}
a-b&=0\\
b+c&=0\\
a+2b+3c&=0.
\end{align*}
The coefficient matrix is given by $$
\left[\begin{array}{rrrr}{1} & {-1} & {0} \\ {0} & {1} & {1} \\ {1} & {2} & {3} \end{array}\right] .
$$
The reduced row-echelon form of the above matrix is given by
$$
\left[\begin{array}{rrrr}{1} & {0} & {1} \\ {0} & {1} & {1} \\ {0} & {0} & {0} \end{array}\right] .
$$
Hence, the system has the solution $a=-t$, $b=-t$, $c=t$.
That is, $$(1,0,1)+(-1,1,2)-(0,1,3)=(0,0,0).$$
