If \[\mathop a\limits^ \to = 2i + 2f + 3k,\mathop b\limits^ \to = - i + 2j + k\& \mathop c\limits^ \to = 3i + j\] are such that \[\mathop a\limits^ \to + \lambda \mathop b\limits^ \to \] is perpendicular to \[\mathop c\limits^ \to \] then find the value of \[\lambda \] .

Hint: It is given that the \[\mathop a\limits^ \to + \lambda \mathop b\limits^ \to \] and \[\mathop c\limits^ \to \] are perpendicular to each other which means that the dot product of both the vectors will be zero.

Complete step by step answer:
As \[\mathop A\limits^ \to \& \mathop B\limits^ \to \] be any two vectors then there dot product is represented by \[\mathop A\limits^ \to .\mathop B\limits^ \to = \left| A \right|\left| B \right|\cos \theta \]
Where \[\theta \] is the angle between the two vectors
If the two vectors are mutually perpendicular then \[\theta = {90^ \circ }\]
Putting this we will get,
\[\begin{array}{l}
 \Rightarrow \mathop A\limits^ \to .\mathop B\limits^ \to = \left| A \right|\left| B \right|\cos {90^ \circ }\\
\mathop { \Rightarrow A}\limits^ \to .\mathop B\limits^ \to = \left| A \right|\left| B \right| \times 0\\
 \Rightarrow \mathop A\limits^ \to .\mathop B\limits^ \to = 0
\end{array}\]
Using this result we can get
As the vectors are \[\mathop a\limits^ \to = 2i + 2f + 3k,\mathop b\limits^ \to = - i + 2j + k\& \mathop c\limits^ \to = 3i + j\]
Then \[\mathop a\limits^ \to + \lambda \mathop b\limits^ \to \] is
\[\begin{array}{l}
\mathop {\therefore {\rm{ }}a}\limits^ \to + \lambda \mathop {{\rm{ }}b}\limits^ \to = (2i + 2j + 3k) + \lambda ( - i + 2j + k)\\
 \Rightarrow \mathop {{\rm{ }}a}\limits^ \to + \lambda \mathop {{\rm{ }}b}\limits^ \to = (2 - \lambda )i + (2 + 2\lambda )j + (3 + \lambda )k
\end{array}\]
Then
\[\begin{array}{l}
\therefore \left( {\mathop {{\rm{ }}a}\limits^ \to + \lambda \mathop {{\rm{ }}b}\limits^ \to } \right).\mathop c\limits^ \to = 0\\
 \Rightarrow \left[ {(2 - \lambda )i + (2 + 2\lambda )j + (3 + \lambda )k} \right].\left[ {3i + j} \right] = 0\\
 \Rightarrow (2 - \lambda )3 + (2 + 2\lambda )1 + (3 + \lambda ) = 0\\
 \Rightarrow 6 - 3\lambda + 2 + 2\lambda = 0\\
 \Rightarrow - \lambda + 8 = 0\\
 \Rightarrow \lambda = 8
\end{array}\]
Which means that the value of \[\lambda \] is 8

Note:
While doing dot product of vectors we must keep in mind that the units of i will only multiply with the units i and the same goes for the rest as it has be obtained that when we do dot product of any other component with i we only get 0 and the same goes for the rest of the components.