If $ \overrightarrow a = \hat i + \hat j - \hat k,\overrightarrow b = \hat i - \hat j + \hat k $ and $ \overrightarrow c $ is a unit vector perpendicular to the vector $ \overrightarrow a $ and coplanar with $ \overrightarrow a $ and $ \overrightarrow b $ , then a unit vector $ \overrightarrow d $ perpendicular to both $ \overrightarrow a $ and $ \overrightarrow c $ is:
 $
  A.\dfrac{1}{{\sqrt 6 }}(2\hat i - \hat j + \hat k) \\
  B.\dfrac{1}{{\sqrt 2 }}(\hat i + \hat j) \\
  C.\dfrac{1}{{\sqrt 2 }}(\hat j + \hat k) \\
  D.\dfrac{1}{{\sqrt 2 }}(\hat i + \hat k) \\
 $

Hint: A vector may be described as a quantity having both magnitude and direction and obeying commutative law of addition. Graphically a vector is represented by a line segment. The direction is indicated by an arrow and the magnitude by the length of the segment.

Complete step-by-step answer:
We have been given three vectors and we have to calculate a unit vector perpendicular to $ \overrightarrow a $ and $ \overrightarrow c $ .
A unit vector is a vector whose magnitude is unity. Also we have been given that $ \overrightarrow c $ is a unit vector perpendicular to vector a and coplanar to $ \overrightarrow a $ and $ \overrightarrow b $ that is all these vectors must be parallel to the same plane. It can be calculated as:
 $
  \Rightarrow \overrightarrow a = \hat i + \hat j - \hat k \\
  \Rightarrow \overrightarrow b = \hat i - \hat j + \hat k \;
 $
Scalar product of two perpendicular vectors is zero, therefore
 $
  \Rightarrow \overrightarrow c \bot \overrightarrow a \Rightarrow \overrightarrow a .\overrightarrow c = 0 \\
  \Rightarrow \left| {\overrightarrow c } \right| = 1 \;
 $
 $ \overrightarrow c $ is coplanar with $ \overrightarrow a $ and $ \overrightarrow b $ and it can be represented as
 $ \Rightarrow \overrightarrow c = \overrightarrow a + \lambda \overrightarrow b $
Multiplying with $ \overrightarrow a $ on both the sides we get,
 $
  \Rightarrow \overrightarrow a .\overrightarrow c = \overrightarrow a .(\overrightarrow a + \lambda \overrightarrow b ) \\
  \therefore \overrightarrow a .\overrightarrow c = 0 \\
  \Rightarrow 0 = {\left| {\overrightarrow a } \right|^2} + \lambda \overrightarrow a .\overrightarrow b \\
  \therefore {\left| {\overrightarrow a } \right|^2} = \sqrt {{1^2} + {1^2} + {{( - 1)}^2}} = \sqrt 3 \\
  \therefore \overrightarrow a .\overrightarrow b = 1 - 1 - 1 = - 1 \\
 $
Substituting the values we get,
 $
  \Rightarrow 0 = \sqrt 3 + \lambda ( - 1) \\
  \Rightarrow \lambda = \sqrt 3 \;
 $
Now vector d is perpendicular to both $ \overrightarrow a $ and $ \overrightarrow c $ .
\[
  \Rightarrow \overrightarrow d = \overrightarrow a \times \overrightarrow c = \overrightarrow a \times (\overrightarrow a + \lambda \overrightarrow b ) \\
  \Rightarrow \overrightarrow d = \overrightarrow a \times \overrightarrow a + \lambda \overrightarrow a \times \overrightarrow b \\
  \Rightarrow \overrightarrow d = 0 + \lambda \overrightarrow a \times \overrightarrow b = \sqrt 3 (\overrightarrow a \times \overrightarrow b ) \\
  \Rightarrow \overrightarrow d = \sqrt 3 (\overrightarrow a \times \overrightarrow {b)} \;
\]
Where \[(\overrightarrow a \times \overrightarrow {b)} \] is can be calculated the following way :
 $
  \Rightarrow (\overrightarrow a \times \overrightarrow {b)} = \left( {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  1&1&{ - 1} \\
  1&{ - 1}&1
\end{array}} \right) = (1 - ( 1))\hat i - \hat j(1 - ( - 1)) + \hat k( - 1 - 1) \\
  \Rightarrow (\overrightarrow a \times \overrightarrow {b)} = 0.\hat i - 2\hat j - 2\hat k \\
 $
Now
 $
  \hat d = \dfrac{{\overrightarrow d }}{{\left| {\overrightarrow d } \right|}} = \dfrac{{\sqrt 3 ( - 2\hat j - 2\hat k)}}{{\left| {\sqrt 3 } \right|\sqrt {{{( - 2)}^2} + {{( - 2)}^2}} }} = \dfrac{{( - 2\hat j - 2\hat k)}}{{2\sqrt 2 }} \\
  \hat d = \dfrac{{ - 1}}{{\sqrt 2 }}(\hat j + \hat k) \;
 $
Hence we can conclude that the correct option is option C.
As we can write the calculated unit vector d as
 $ \hat d = \pm \dfrac{1}{{\sqrt 2 }}(\hat j + \hat k) $
So, the correct answer is “Option C”.

Note: We can multiply the vectors by two methods: scalar product or dot product and vector product or cross product. Both of them have the properties of their own. Unlike scalar product, vector product is neither associative nor commutative. It can be calculated as;
 $ \Rightarrow (\overrightarrow a \times \overrightarrow {b)} = \left( {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  {{a_1}}&{{a_2}}&{{a_3}} \\
  {{b_1}}&{{b_2}}&{{b_3}}
\end{array}} \right) $