Let a → = i ^ - j ^ ,   b → = j ^ - k ^ and c → = k ^ - i ^ . If d → is a unit…

Let $\overrightarrow{d}=d_1\hat{i}+d_2\hat{j}+d_3\hat{k}$.

Given $\overrightarrow{a}\cdot \overrightarrow{d}=0$.

$\Rightarrow d_1-d_2=0$

$\Rightarrow d_1=d_2 ...........\left(\mathrm{i}\right)$

Also, given $\overrightarrow{d}$ is a unit vector.

$\Rightarrow {d_1}^2+{d_2}^2+{d_3}^2=1 .........\left(\mathrm{ii}\right)$

Also, given $\left[\begin{array}{ccc}\overrightarrow{b}& \overrightarrow{c}& \overrightarrow{d}\end{array}\right]=0$.

$\Rightarrow \left|\begin{array}{ccc}0& 1& -1\\ -1& 0& 1\\ d_1& d_2& d_3\end{array}\right|=0$

$\Rightarrow -1\left(-d_3-d_1\right)-1\left(-d_2\right)=0$

$\Rightarrow d_1+d_2+d_3=0$

$\Rightarrow 2d_1+d_3=0$ (from equation $\left(\mathrm{i}\right)$)

$\Rightarrow d_3=-2d_1 ...........\left(\mathrm{iii}\right)$

Putting equations $\left(\mathrm{i}\right) \& \left(\mathrm{iii}\right)$ in equation $\left(\mathrm{ii}\right)$, we get

${d_1}^2+{d_1}^2+4{d_1}^2=1$

$\Rightarrow d_1=\pm \frac{1}{\sqrt{6}}$

$\therefore d_2=\pm \frac{1}{\sqrt{6}}$ and $d_3=\mp \frac{2}{\sqrt{6}}$.

Hence, the required vector is $\overrightarrow{d}=\pm \frac{\hat{i}+\hat{j}-2\hat{k}}{\sqrt{6}}$.