If the variable line, $3x+4y-\lambda =0$ is such that the two circles ${{x}^{2}}+{{y}^{2}}-2x-2y+1=0$ and ${{x}^{2}}+{{y}^{2}}-18x-2y+78=0$ are on its opposite sides, then the set of all values of $\lambda $in the interval:
(a) [12, 21]
(b) (2, 17)
(c) (23, 31)
(d) [13, 23]

Hint: First, by using the completing the square method for both circles, find their center and radius. Then, by using the above condition and also the equation of line $3x+4y-\lambda =0$and by substituting the value of centre for circle${{C}_{1}}$ which is $\left( 1,1 \right)$and ${{C}_{2}}$ with centre $\left( 9,1 \right)$, we get the belonging range of $\lambda $. Then, by using the distance formula for the circle from the line equation given, we get the range of values of $\lambda $.

Complete step by step answer:
In this question, we are supposed to find the set of all values of $\lambda $in the interval when the variable line, $3x+4y-\lambda =0$is such that the two circles ${{x}^{2}}+{{y}^{2}}-2x-2y+1=0$and ${{x}^{2}}+{{y}^{2}}-18x-2y+78=0$are on its opposite sides.
So, by using completing the square for the first circle equation as ${{x}^{2}}+{{y}^{2}}-2x-2y+1=0$ and adding and subtracting 1 on the left hand side, we get:
$\begin{align}
  & {{x}^{2}}+{{y}^{2}}-2x-2y+1-1+1=0 \\
 & {{\left( x-1 \right)}^{2}}+{{\left( y-1 \right)}^{2}}={{1}^{2}} \\
\end{align}$
So, we get the circle with centre $\left( 1,1 \right)$and radius 1.
Similarly, by using completing the square for the first circle equation as ${{x}^{2}}+{{y}^{2}}-2x-2y+1=0$ and adding and subtracting 1 and 81 on the left hand side, we get:
$\begin{align}
  & {{x}^{2}}+{{y}^{2}}-2x-2y+1-1+1-81+81=0 \\
 & {{\left( x-9 \right)}^{2}}+{{\left( y-1 \right)}^{2}}={{2}^{2}} \\
\end{align}$
So, we get the circle with centre $\left( 9,1 \right)$and radius 2.
Now, we are given in the question that circles ${{C}_{1}}$and ${{C}_{2}}$ are on the opposite sides which gives the condition as:
${{C}_{1}}{{C}_{2}}<0$
So, by using the above condition and also the equation of line $3x+4y-\lambda =0$and by substituting the value of centre for circle${{C}_{1}}$which is $\left( 1,1 \right)$and ${{C}_{2}}$with centre $\left( 9,1 \right)$, we get:
$\left( 3\times 1+4\times 1-\lambda \right)\left( 3\times 9+4\times 1-\lambda \right)<0$
Then, by solving the above expression, we get:
\[\begin{align}
  & \left( 3+4-\lambda \right)\left( 27+4-\lambda \right)<0 \\
 & \Rightarrow \left( 7-\lambda \right)\left( 31-\lambda \right)<0 \\
\end{align}\]
So, we get range of values of $\lambda $ from the conditions as:
$\begin{align}
  & 7-\lambda <0 \\
 & \Rightarrow 7<\lambda \\
\end{align}$ and $\begin{align}
  & 31-\lambda <0 \\
 & \Rightarrow 31<\lambda \\
\end{align}$
So, we get the condition that $\lambda \in \left( 7,31 \right)$.
Then, by using the distance formula for the circle${{C}_{1}}$for the above condition of $\lambda $, we get:
$\left| \dfrac{3+4-\lambda }{\sqrt{{{3}^{2}}+{{4}^{2}}}} \right|\ge 1$
So, by solving the above expression, we get:
$\begin{align}
  & \left| \dfrac{7-\lambda }{\sqrt{9+16}} \right|\ge 1 \\
 & \Rightarrow \left| \dfrac{7-\lambda }{\sqrt{25}} \right|\ge 1 \\
 & \Rightarrow \left| \dfrac{7-\lambda }{5} \right|\ge 1 \\
 & \Rightarrow \left| 7-\lambda \right|\ge 5 \\
\end{align}$
So, we get two values of $\lambda $from the above expression as:
$\begin{align}
  & 7-\lambda =5 \\
 & \Rightarrow \lambda =2 \\
\end{align}$ and $\begin{align}
  & 7-\lambda =-5 \\
 & \Rightarrow \lambda =12 \\
\end{align}$
So, we can see clearly that the range $\left( -\infty ,2 \right]\cup \left[ 12,\infty \right)$satisfies the existence of $\lambda $.
Similarly, by using the distance formula for the circle${{C}_{2}}$for the above condition of $\lambda $, we get:
$\left| \dfrac{27+4-\lambda }{\sqrt{{{3}^{2}}+{{4}^{2}}}} \right|\ge 2$
So, by solving the above expression, we get:
$\begin{align}
  & \left| \dfrac{31-\lambda }{\sqrt{9+16}} \right|\ge 2 \\
 & \Rightarrow \left| \dfrac{31-\lambda }{\sqrt{25}} \right|\ge 2 \\
 & \Rightarrow \left| \dfrac{31-\lambda }{5} \right|\ge 2 \\
 & \Rightarrow \left| 31-\lambda \right|\ge 10 \\
\end{align}$
So, we get two values of $\lambda $from the above expression as:
$\begin{align}
  & 31-\lambda =10 \\
 & \Rightarrow \lambda =21 \\
\end{align}$ and $\begin{align}
  & 31-\lambda =-10 \\
 & \Rightarrow \lambda =41 \\
\end{align}$
So, we can see clearly that the range $\left( -\infty ,21 \right]\cup \left[ 41,\infty \right)$satisfies the existence of $\lambda $.
So, we get the actual range of existence of $\lambda $from the above two conclusions is [12, 21].
Hence, option (a) is correct.

Note:
Now, to solve these type of the questions we need to know some of the basic formula for the distance that is used in this question for the circle with the line as $ax+by-\lambda =0$is given by the equation for the point $\left( {{x}_{1}},{{y}_{1}} \right)$as:
$\left| \dfrac{a{{x}_{1}}+b{{y}_{1}}-\lambda }{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|\ge r$ where r is the radius of the circle given.