A unit vector which is perpendicular to the vector $ 2\widehat i - \widehat j + 2\widehat k $ and is coplanar with the vectors $ \widehat i + \widehat j - \widehat k $ and $ 2\widehat i + 2\widehat j - \widehat k $ is
A. $ \dfrac{{2\widehat j + \widehat k}}{{\sqrt 5 }} $
B. $ \dfrac{{3\widehat i + 2\widehat j - 2\widehat k}}{{\sqrt {17} }} $
C. $ \dfrac{{3\widehat i + 2\widehat j + 2\widehat k}}{{\sqrt {17} }} $
D. $ \dfrac{{2\widehat i + 2\widehat j - \widehat k}}{3} $

Hint: The unit vector is also known as the normalized vector. Here we will first suppose the unique vector as the sum of the two given vectors. Then find the values for the unit vector and then place the values in the standard formula for the resultant value.

Complete step-by-step answer:
Let us assume that the unit vector be $ x\widehat i + y\widehat j + j\widehat k $
Unit vector can be expressed as
 $ \sqrt {{x^2} + {y^2} + {j^2}} = 1 $
Squaring both the sides of the equation –
 $ \Rightarrow {x^2} + {y^2} + {j^2} = 1 $ .... (A)
A unit vector which is perpendicular to the vector $ 2\widehat i - \widehat j + 2\widehat k $ and is coplanar with the vectors $ \widehat i + \widehat j - \widehat k $ and $ 2\widehat i + 2\widehat j - \widehat k $
 $ 2x - y + 2j = 0 $ .... (B)
 $ \left| {\begin{array}{*{20}{c}}
  x&y&z \\
  1&1&{ - 1} \\
  2&1&{ - 1}
\end{array}} \right| $
Expand the above determinant –
 $ = x( - 1 + 1) - y( - 2 + 1) + z(1 - 2) $
Simplify the above equation –
 $ \Rightarrow y = - j $
Place the above value in equation (B)
 $ 2x + 3j = 0 $
Make “X” the subject-
 $ 2x = - 3j $
When the term multiplicative on one side is moved to the opposite side then it goes to denominator.
 $ \Rightarrow x = \dfrac{{ - 3j}}{2} $
Place in equation (A)
 $ \dfrac{{9{j^2}}}{4} + {j^2} + {j^2} = 1 $
Take LCM (Least common multiple) and simplify the above equation-
 $ \Rightarrow \dfrac{{9{j^2}}}{4} + \dfrac{{4{j^2}}}{4} + \dfrac{{4{j^2}}}{4} = 1 $
When denominators are the same, numerators are added.
 $
   \Rightarrow \dfrac{{9{j^2} + 4{j^2} + 4{j^2}}}{4} = 1 \\
   \Rightarrow \dfrac{{17{j^2}}}{4} = 1 \;
  $
Do cross multiplication and simplify the above equation –
 $
   \Rightarrow 17{j^2} = 4 \\
   \Rightarrow {j^2} = \dfrac{4}{{17}} \;
  $
Take square-root on both the sides –
 $ \Rightarrow \sqrt {{j^2}} = \sqrt {\dfrac{4}{{17}}} $
Square and square-root cancel each other on left hand side of the equation-
 $ \Rightarrow j = \pm \dfrac{2}{{\sqrt {17} }} $
So, the required vector is – $ \dfrac{{3\widehat i + 2\widehat j - 2\widehat k}}{{\sqrt {17} }} $
So, the correct answer is “Option B”.

Note: Be good in square and square-root and simplify the resultant value accordingly. Always remember that the square of negative number of positive number always gives positive value. Coplanar vectors are the vectors which lie in the three-dimensional space on the same plane