If \[f(x)=\left| \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{3}} & 1 \\
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{3}} & 1 \\
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{3}} & 1 \\
\end{matrix} \right|\], then \[f'(x)=\lambda \left| \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{2}} & 1 \\
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{2}} & 1 \\
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{2}} & 1 \\
\end{matrix} \right|\]. Find the value of \[\lambda .\]

Hint: Take the matrix f(x) and find its determinant f’(x), i.e. differentiate \[{{C}_{1}}\] while \[{{C}_{2}}\] and \[{{C}_{3}}\] are constant plus differentiate \[{{C}_{2}}\] while \[{{C}_{1}}\] and \[{{C}_{3}}\] are constant. Thus, the columns become some value, apply property of determinants and solve it.

Complete step-by-step answer:
Given to us is that \[f(x)=\left| \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{3}} & 1 \\
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{3}} & 1 \\
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{3}} & 1 \\
\end{matrix} \right|\] and \[f'(x)=\lambda \left| \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{2}} & 1 \\
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{2}} & 1 \\
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{2}} & 1 \\
\end{matrix} \right|........(1)\]
In the matrix of \[f(x)=\left| \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{3}} & 1 \\
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{3}} & 1 \\
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{3}} & 1 \\
\end{matrix} \right|\]
Here, \[{{C}_{1}}\], \[{{C}_{2}}\] and \[{{C}_{3}}\] are the three columns of the \[3\times 3\] matrix.
Now let us differentiate column 1 and keep \[{{C}_{2}}\] and \[{{C}_{3}}\] as they are in the first. Then keep \[{{C}_{1}}\] and \[{{C}_{3}}\] constant and differentiate \[{{C}_{2}}\]. If we differentiate \[{{C}_{3}}\], we will get the column as zero.
\[f'(x)=\left| \begin{matrix}
   4{{\left( x-a \right)}^{3}} & {{\left( x-a \right)}^{3}} & 1 \\
   4{{\left( x-b \right)}^{3}} & {{\left( x-b \right)}^{3}} & 1 \\
   4{{\left( x-c \right)}^{3}} & {{\left( x-c \right)}^{3}} & 1 \\
\end{matrix} \right|+\left| \begin{matrix}
   {{\left( x-a \right)}^{4}} & 3{{\left( x-a \right)}^{2}} & 1 \\
   {{\left( x-b \right)}^{4}} & 3{{\left( x-b \right)}^{2}} & 1 \\
   {{\left( x-c \right)}^{4}} & 3{{\left( x-c \right)}^{2}} & 1 \\
\end{matrix} \right|\]
\[=4\left| \begin{matrix}
   {{\left( x-a \right)}^{3}} & {{\left( x-a \right)}^{3}} & 1 \\
   {{\left( x-b \right)}^{3}} & {{\left( x-b \right)}^{3}} & 1 \\
   {{\left( x-c \right)}^{3}} & {{\left( x-c \right)}^{3}} & 1 \\
\end{matrix} \right|+3\left| \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{2}} & 1 \\
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{2}} & 1 \\
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{2}} & 1 \\
\end{matrix} \right|\]
\[{{C}_{1}}\] and \[{{C}_{2}}\] in the first determinant is same, so the value becomes zero, i.e. \[{{C}_{1}}={{C}_{2}}=0\].
i.e. the determinant of a matrix with two identical columns is zero.
Thus, \[f'(x)=3\left| \begin{matrix}
   {{\left( x-a \right)}^{4}} & {{\left( x-a \right)}^{2}} & 1 \\
   {{\left( x-b \right)}^{4}} & {{\left( x-b \right)}^{2}} & 1 \\
   {{\left( x-c \right)}^{4}} & {{\left( x-c \right)}^{2}} & 1 \\
\end{matrix} \right|......(2)\]
Now let us compare equation (1) and (2).
From this, we can say that \[\lambda =3\].
Thus, we got the value of \[\lambda =3\].

Note: It is one of the properties of determinants. If we have a matrix with two identical rows or columns then its determinant is equal to zero. In this question we had two columns same, i.e. \[{{C}_{1}}={{C}_{2}}\]. Thus, finding its determinant comes out, to be zero at the end.