Let $$\overrightarrow a = \widehat i + \widehat j + 2\widehat k$$, $$\overrightarrow b = \widehat i + 2\widehat j + \widehat k$$ and $$\overrightarrow c = \widehat i + \widehat j + \widehat k$$.

Any vector in the plane of $$\widehat i + \widehat j + 2\widehat k$$ and $$\widehat i + 2\widehat j + \widehat k$$ is given by

$$\overrightarrow r = \lambda \overrightarrow a + \mu \overrightarrow b $$

$$ = \lambda (\widehat i + \widehat j + 2\widehat k) + \mu (\widehat i + 2\widehat j + \widehat k)$$

$$ = (\lambda + \mu )\widehat i + (\lambda + 2\mu )\widehat j + (2\lambda + \mu )\widehat k$$

Also, $$\overrightarrow r \,.\,\overrightarrow c = 0$$

$$ \Rightarrow (\lambda + \mu )\,.\,1 + (\lambda + 2\mu )\,.\,1 + (2\lambda + \mu )\,.\,1 = 0$$

$$ \Rightarrow 4\lambda + 4\mu = 0$$

$$ \Rightarrow \lambda + \mu = 0$$

$$ \Rightarrow \left[ {\matrix{ {\overrightarrow r } & {\overrightarrow a } & {\overrightarrow b } \cr } } \right] = 0$$

So, vectors $$\widehat j - \widehat k$$ and $$ - \widehat j + \widehat k$$ satisfy this.