If the vector $a = 2i + 3j + 6k$ and b are collinear and $\left| b \right| = 21,$ then b ¾
A.$ \pm \left( {2i + 3j + 6k} \right)$
B.$ \pm \;3\left( {2i + 3j + 6k} \right)$
C.$\left( {2i + j + k} \right)$
D.$ \pm 21\left( {2i + 3j + 6k} \right)$

Hint: The vector being collinear means it lines on the same line or parallel line, might having different length. The Magnitude of vector may varies and can be calculated for $z = xi + yj\;\;as\left| z \right| = \sqrt {{x^2} + {y^2}} $

Complete step-by-step answer:
The given vector $a = 2i + 3j + 6k$ has another vector b collinear which means we have n as a vector in the same unit direction of a or opposite direction.
First we will find the unit vector in the direction of vector of vector a, which is equals to
\[ \Rightarrow \;\hat a\; = \;\dfrac{{\vec a}}{{\left| {\vec a} \right|}}\]
So \[\left| {\vec a} \right|\; = \;\sqrt {{2^2} + {3^2} + {6^2}} = \sqrt {4 + 9 + 36} = \sqrt {49} = 7\]
We get the unit vector as,
\[\hat a\; = \;\dfrac{{\left( {2i + 3j + 6k} \right)}}{7} = \dfrac{1}{7}\;\left( {2i + 3j + 6k} \right)\]
The b vector being collinear ,means can be in same or opposite direction so the unit vector will be \[ \pm \,\hat a\]
Hence, unit vector for b is
$ \Rightarrow \hat b = \pm \hat a = \pm \dfrac{1}{7}\left( {2i + 3j + 6k} \right)$
As the given information that Magnitude of $\vec b$ is 21 as we know that
     $\vec b = \left| {\vec b} \right|\;\hat b$
So the resultant vector of b will comes out to be,
     $ \Rightarrow \vec b = \pm 21 \times \dfrac{1}{7}\left( {2i + 3j + 6k} \right)$
$ \Rightarrow \vec b = \pm 3\left( {2i + 3j + 6k} \right)$

Thus, option B is the correct Answer.

Note: The chance of mistake is almost in direction. Collinear means the unit vector in the same direction or exact opposite direction. It also means they are co-director in the same line.