If $ \overrightarrow r = \left( {x + y + 2} \right)\widehat i + \left( {2x - y + 3} \right)\widehat j + \left( {x + 2y + 7} \right)\widehat k $
Where \[\overrightarrow r .\widehat i = 3,\overrightarrow r .\widehat j = 5\] then $ \overline r .\widehat k = $
(A) $ 4 $
(B) $ 6 $
(C) $ 9 $
(D) $ 8 $

Hint: This type of question will be solved by using dot product properties. Remember the vector multiplication is nowhere similar to that of scalar multiplication as it involves direction along with magnitude.

Complete step-by-step answer:
In this question given that
 $ \overrightarrow r = \left( {x + y + 2} \right)\widehat i + \left( {2x - y + 3} \right)\widehat j + \left( {x + 2y + 7} \right)\widehat k $ . . . (1)
and
\[\overrightarrow r .\widehat i = 3\]
 $ \overrightarrow r .\widehat j = 5 $
 $ \overrightarrow r .\widehat k = ? $
Multiply equation (1) by $ \widehat i $
\[ \Rightarrow \overrightarrow r .\widehat i = \left( {x + y + 2} \right)\widehat i.\widehat i + \left( {2x - y + 3} \right)\widehat j.\widehat i + \left( {x + 2y + 7} \right)\widehat k.\widehat i\] . . . (2)
 $ \widehat i.\widehat i = \left| {\widehat i} \right|\left| {\widehat i} \right|\cos \theta = 1 $ $ (\because \theta = 0 $ and $ \cos 0 = 1) $
 $ \widehat i.\widehat j = \left| {\widehat i} \right|\left| {\widehat j} \right|\cos \theta = 0 $ $ (\because \theta = {90^0} $ and $ \cos {90^0} = 0) $
Similarly, $ \widehat i.\widehat k = 0 $
By the above rule we can simplify equation (2) as,
 $ \Rightarrow 3 = x + y + 2 + 0 + 0 $
On simplifying we get the
 $ x + y = 1 $ . . . . (3)
Now,
Multiply equation (1) by $ \widehat j $
\[\overrightarrow r .\widehat j = \left( {x + y + 2} \right)\widehat i.\widehat j + \left( {2x - y + 3} \right)\widehat j.\widehat j + \left( {x + 2y + 7} \right)\widehat k.\widehat j\]
From the explanation we gave above, we can write
\[\widehat i.\widehat j = \widehat j.\widehat k = 0\] and \[\widehat j.\widehat j = 1\]
 $ \Rightarrow 5 = 0 + 2x - y + 3 + 0 $
On simplifying the above equation, we get
 $ 2x - y + 3 = 5 $
 $ \Rightarrow 2x - y = 2 $ . . . (4)
Now, multiply equation (1) by \[\widehat k\]
\[\overrightarrow r .\widehat k = \left( {x + y + 2} \right)\widehat i.\widehat k + \left( {2x - y + 3} \right)\widehat j.\widehat k + \left( {x + 2y + 7} \right)\widehat k.\widehat k\]
Again, From the explanation we gave above, we can write
\[\widehat i.\widehat k = \widehat j.\widehat k = 0\] and \[\widehat k.\widehat k = 1\]
 $ \Rightarrow \overrightarrow r .\widehat k = 0 + 0 + x + 2y + 7 $
 $ \Rightarrow \overrightarrow r .\widehat k = x + 2y + 7 $ . . . (5)
Adding equation (3) and (4), we get
 $ 3x = 3 $
 $ \Rightarrow x = 1 $
Put this value of $ x $ in equation (3)
 $ \Rightarrow 1 + y = 1 $
 $ \Rightarrow y = 0 $
By substituting these values of $ x $ and $ y $ in equation (5) we get
 $ \Rightarrow \overrightarrow r .\widehat k = 1 + 2 \times 0 + 7 $
 $ \Rightarrow \overrightarrow r .\widehat k = 1 + 7 $
 $ \Rightarrow \overrightarrow r .\widehat k = 8 $
Therefore, from the above explanation, the correct answer is option (D) $ 8 $

So, the correct answer is “Option D”.

Note: Unit vectors $ \widehat i,\widehat j $ and $ \widehat k $ represent positive X, Y and Z axis, respectively. Therefore, they are perpendicular to each other. Dot product of two perpendicular vectors is zero, since the angle between them is $ {90^0} $ and $ \cos {90^0} = 0 $ .
Perpendicular vectors are also called orthogonal vectors.