Let three vectors \[\overrightarrow{a}\] , \[\overrightarrow{b}\] and \[\overrightarrow{c}\] be such that \[\overrightarrow{c}\] is coplanar with \[\overrightarrow{a}\] and \[\overrightarrow{b}\] , \[\overrightarrow{a}.\overrightarrow{c} = 7\] and \[\overrightarrow{b}\] is perpendicular to \[\overrightarrow{c}\] , where \[\overrightarrow{a} = - \hat{i} + \hat{j} + \hat{k}\] and \[\overrightarrow{b} = 2\ \hat{i} + \hat{k}\] . Then the value of \[2 \mid \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \mid^{2}\] is

Hint: In this question, we need to find the value of \[2 \mid \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \mid^{2}\] . First we need to use the given conditions \[\overrightarrow{c}\] is coplanar with \[\overrightarrow{a}\] and \[\overrightarrow{b}\] and \[\overrightarrow{b}\] is perpendicular to \[\overrightarrow{c}\] . Next we need to find the value of \[\overrightarrow{c}\] with the help of \[\overrightarrow{a}\] and \[\overrightarrow{b}\] . Then substituting the values of \[\overrightarrow{a}\] , \[\overrightarrow{b}\ \] and \[\overrightarrow{c}\] . We can find the value of \[2 \mid \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \mid^{2}\]

Complete step by step Solution:
Given that \[\overrightarrow{c}\] is coplanar with \[\overrightarrow{a}\] and \[\overrightarrow{b}\] and \[\overrightarrow{b}\] is perpendicular to \[\overrightarrow{c}\]
That is,
\[\overrightarrow{c} = \lambda(\overrightarrow{b}(\overrightarrow{a} \times \overrightarrow{b})\]
\[\overrightarrow{c} = \lambda\left( \left( \overrightarrow{b}.\overrightarrow{b} \right)\overrightarrow{a} - \left( \overrightarrow{\text{b.}}\overrightarrow{a} \right)\overrightarrow{b} \right)\] •••• (1)
First, we can find \[\overrightarrow{b}.\overrightarrow{b}\]
Given that \[\overrightarrow{b} = 2\ \hat{i} + \hat{k}\]
Now
\[\overrightarrow{b}.\overrightarrow{b} = (2\ \hat{i} + \hat{k}).(\ 2\ \hat{i} + \hat{k})\]
On solving,
We get,
\[\overrightarrow{b}.\overrightarrow{b} = \left( 2 \times 2 \right) + \left( 1 \times 1 \right)\]
On simplifying,
We get,
\[\overrightarrow{b}.\overrightarrow{b} = 5\]
Similarly we need to find \[\overrightarrow{b}.\overrightarrow{a}\]
Given that \[\overrightarrow{a} = - \hat{i} + \hat{j} + \hat{k}\] and \[\overrightarrow{b} = 2\ \hat{i} + \hat{k}\]
Now,
\[\overrightarrow{b}.\overrightarrow{a} = \left( 2\ \hat{i} + \hat{k} \right).\left( - \hat{i} + \hat{j} + \hat{k} \right)\]
On simplifying,
We get,
\[\overrightarrow{b}.\overrightarrow{a} = \left( 2 \times – 1) – (0 \times 1 \right) + (1 \times 1)\]
On further simplification,
We get,
\[\overrightarrow{b}.\overrightarrow{a} = - 2 + 1 = - 1\]
Now on substituting the values of \[\overrightarrow{b}.\overrightarrow{b}\] and \[\overrightarrow{b}.\overrightarrow{a}\] in (1) ,
We get,
\[\overrightarrow{c} = \lambda\left( 5(\overrightarrow{a}) – ( - 1)(\overrightarrow{b}) \right)\]
On substituting the value of \[\overrightarrow{a}\] and \[\overrightarrow{b}\] ,
We get,
\[\overrightarrow{c} = \lambda\left( 5\left( - \hat{i} + \hat{j} + \hat{k} \right) + (2\ \hat{i} + \hat{k}) \right)\]
On solving,
We get,
\[\overrightarrow{c} = \lambda\left( \left( - 5\hat{i} + 5\hat{j} + 5\hat{k} \right) + (2\ \hat{i} + \hat{k}) \right)\]
Again on solving,
We get,
\[\overrightarrow{c} = \lambda\left( \left( - 3\hat{i} + 5\hat{j} + 6\hat{k} \right) \right)\] ••• (2)
Also given that \[\overrightarrow{a}.\overrightarrow{c} = 7\]
On substituting \[\overrightarrow{a}\] and \[\overrightarrow{c}\],
We get,
\[( - \hat{i} + \hat{j} + \hat{k}).\lambda\left( - 3\hat{i} + 5\hat{j} + 6\hat{k} \right) = 7\]
On simplifying,
We get,
\[( - 1 \times – 3\lambda) + (1 \times 5\lambda) + (1 \times 6\lambda) = 7\]
On further simplification,
We get,
\[3\lambda + 5\lambda + 6\lambda = 7\]
In adding all,
We get,
\[14\lambda = 7\]
On dividing both sides by \[14\] ,
We get,
\[\lambda = \dfrac{1}{2}\]
Now on substituting the value of \[\lambda\] in \[\overrightarrow{c} = \lambda\left( - 3\hat{i} + 5\hat{j} + 6\hat{k} \right)\]
We get,
\[\overrightarrow{c} = \dfrac{1}{2}\left( - 3\hat{i} + 5\hat{j} + 6\hat{k} \right)\ \]
We need to find the value of \[2 \mid \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \mid^{2}\]
On substituting all the values,
We get,
\[2 \mid \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \mid^{2} = 2\left( \left| \left( - \hat{i} + \hat{j} + \hat{k} \right) + \left( 2\hat{i} + \hat{k} \right) + \dfrac{1}{2}\left( - 3\hat{i} + 5\hat{j} + 6\hat{k} \right) \right| \right)^{2}\]
On simplifying,
We get,
\[2 \mid \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \mid^{2} = 2\left( \left| \left( - 1 + 2 - \dfrac{3}{2} \right)\hat{i} + \left( 1 + 0 + \dfrac{5}{2} \right)\hat{j} + \left( 1 + 1 + 3 \right)\hat{k} \right| \right)^{2}\]
On further simplification,
We get,
\[2 \mid \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \mid^{2} = 2\left( \left| \left( \dfrac{- 2 + 4 – 3}{2} \right)\hat{\text{i\ }} + \left( \dfrac{2 + 5}{2} \right)\hat{j} + \left( 1 + 1 + 3 \right)\hat{k} \right| \right)^{2}\]
On solving,
We get,
\[2 \mid \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \mid^{2} = 2\left( \left| \left( \dfrac{1}{2} \right)\hat{\text{i\ }} + \left( \dfrac{7}{2} \right)\hat{j} + \left( 5 \right)\hat{k} \right| \right)^{2}\ \]
On further solving,
\[2 \mid \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \mid^{2} = 2\left( \dfrac{1}{4} + \dfrac{49}{4} + 25 \right)\]
On taking LCM,
We get,
\[2 \mid \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \mid^{2} = 2\left( \dfrac{1 + 49 + 100}{4} \right)\]
On simplifying,
We get,
\[2 \mid \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \mid^{2} = 2\left( \dfrac{150}{4} \right)\]
On further simplification,
We get,
\[2 \mid \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \mid^{2} = 75\]
Thus the value of \[2 \mid \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \mid^{2}\] is \[75\]

Answer :
The value of \[2 \mid \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} \mid^{2}\] is \[75\].

Note:In order to solve these types of questions, we should have a strong grip over vectors and coplanar vectors. Also, we should know about the product of vectors, there are two types of products in vectors. They are dot products and cross products. Two vectors’ dot product is a scalar that is located in the plane of the two vectors. A vector that is perpendicular to the plane in which these two vectors are located is the result of two vectors being cross-products.