Let

$\overrightarrow{b}=x\hat{i}+y\hat{j}+z\hat{k}$

Then,

$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1& -1& 1\\ x& y& z\end{array}\right|$

$\Rightarrow \overrightarrow{a}\times \overrightarrow{b}=-\left(y+z\right)\hat{i}+\left(x+z\right)\hat{j}+\left(x-y\right)\hat{k}$

So,

$-\left(y+z\right)\hat{i}+\left(x+z\right)\hat{j}+\left(x-y\right)\hat{k}=\hat{i}-\hat{j}+0\hat{k}$

Hence,

$$\begin{gathered} y+z=-1 \\ x+z=-1 \\ x-y=0\Rightarrow x=y \end{gathered}$$

Also,

$\overrightarrow{a}·\overrightarrow{b}=1$

$\Rightarrow -x-y+z=1$

$\Rightarrow -2x+z=1$

$\Rightarrow z=2x+1$

$\Rightarrow -1-x=2x+1$

$\Rightarrow 3x=-2\Rightarrow x=-\frac{2}{3}$

So, $y=-\frac{2}{3}; z=-\frac{1}{3}$

Hence,

$\overrightarrow{b}=-\frac{2}{3}\hat{i}-\frac{2}{3}\hat{j}-\frac{1}{3}\hat{k}$

$\Rightarrow 6\overrightarrow{b}=-4\hat{i}-4\hat{j}-2\hat{k}$

So,

$\overrightarrow{a}-6\overrightarrow{b}=3\left(\hat{i}+\hat{j}+\hat{k}\right)$