Write a unit vector perpendicular to both the vectors $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}==\hat{i}+\hat{j}$.

Hint: Now we know that We know the perpendicular vector of two vectors is given by cross product of two vectors. Cross product of the two vectors ${{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}$ and ${{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}$ is given by determinant of matrix $\left[ \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
\end{matrix} \right]$ Hence we can easily find perpendicular to both the vectors $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}==\hat{i}+\hat{j}$.
Now once we have the vector perpendicular to both the vectors $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}==\hat{i}+\hat{j}$we will divide the vector by its modulus hence we get a unit vector.

Complete step-by-step answer:
Now consider the two given vectors $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}==\hat{i}+\hat{j}$
Any vector perpendicular to ${{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}$ and ${{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}$ is given by $({{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k})\times ({{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k})$and that is the cross product of two vectors
And the cross product of the two vectors ${{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}$ and ${{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}$ is given by determinant of matrix $\left[ \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
\end{matrix} \right]$.
Hence the vector perpendicular to ${{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}$ and ${{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}$ is given by
$\left| \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   1 & 1 & 1 \\
   1 & 0 & 1 \\
\end{matrix} \right|$
Now let us open the determinant with respect to first row
$\begin{align}
  & \hat{i}[1(1)-0(1)]-\hat{j}[1(1)-1(1)]+\hat{k}[1(0)-(1)(1)] \\
 & =\hat{i}-\hat{k} \\
\end{align}$
Let us call this vector as $\vec{c}$
$\vec{c}=\hat{i}-\hat{k}$
Hence now we have the vector perpendicular to both the vectors $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}==\hat{i}+\hat{j}$ is $\vec{c}=\hat{i}-\hat{k}$
Now since we need a unit vector perpendicular to both the vectors $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}==\hat{i}+\hat{j}$
Hence to make $\vec{c}=\hat{i}-\hat{k}$ we will have to divide the vector by its modulus. Modulus of a vector is the distance of the vector from origin
Now modulus of a vector ${{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}$ is given by the formula $\sqrt{{{a}_{1}}^{2}+{{a}_{2}}^{2}+{{a}_{3}}^{2}}$
Hence using this we get modulus of $\vec{c}=\hat{i}-\hat{k}$ as $|\vec{c}|=\sqrt{{{1}^{2}}+{{(-1)}^{2}}}=\sqrt{2}$
Hence now we divide the vector with its modulus and hence the vector obtained is a unit vector
$\dfrac{{\vec{c}}}{|\vec{c}|}=\dfrac{\hat{i}-\hat{k}}{\sqrt{2}}=\dfrac{1}{\sqrt{2}}\hat{i}-\dfrac{1}{\sqrt{2}}\hat{k}$
Hence we have the unit vector perpendicular to both the vectors $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}==\hat{i}+\hat{j}$is $\dfrac{1}{\sqrt{2}}\hat{i}-\dfrac{1}{\sqrt{2}}\hat{k}$.

Note: Now remember that cross product of ${{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}$ and ${{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}$ is given by determinant of matrix $\left[ \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
\end{matrix} \right]$ not to be confused with dot product which is given by ${{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}$ .