Point $(1,2)$ lies on the circle ${x^2} + {y^2} + 2x + 2y – 11 = 0\,\,\,$, because coordinates of point $(1,2)$ satisfy the equation

${x^2} + {y^2} + 2x + 2y – 11 = 0\,\,\,$

Now, ${x^2} + {y^2} – 4x + 6y – 21 = 0\,\,\,\,\,\,\,\,\,……..\left( i \right)$

${x^2} + {y^2} + 2x + 2y – 11 = 0\,\,\,\,\,\,\,\,\,……….\left( {ii} \right)$

$3x + 4y + 5 = 0\,\,\,\,\,\,\,\,\,……….\left( {iii} \right)$

From $(i)$ and $(iii)$

${x^2} + {\left( { – \frac{{3x + 5}}{4}} \right)^2} – 4x – 6{\left( { – \frac{{3x + 5}}{4}} \right)^2} – 21 = 0$

$\begin{array}{l}
 \Rightarrow 16{x^2} + 9{x^2} + 30x + 25 – 64x\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 72x + 120 – 336 = 0
\end{array}$

$ \Rightarrow 25{x^2} + 38x – 191 = 0\,\,\,\,\,\,\,\,\,\,…….\left( {iv} \right)$

From $(ii)$ abd $(iii)$,

${x^2} + {\left( { – \frac{{3x + 5}}{4}} \right)^2} + 2x + 2{\left( { – \frac{{3x + 5}}{4}} \right)^2} – 11 = 0$

$\begin{array}{l}
 \Rightarrow 16{x^2} + 9{x^2} + 30x + 25 + 32x\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, – 24x – 40 – 176 = 0
\end{array}$

$ \Rightarrow 25{x^2} + 38x – 191 = 0\,\,\,\,\,\,\,\,\,\,…….\left( v \right)$

Thus we get the same equation from $(ii)$ and $(iii)$ as we get from equation $(i)$ and $(iii)$. Hence the point of intersction of $(ii)$ and $(iii)$ will be sameas the point of intersections of $(i)$ and $(iii)$, Therefore the circle $(ii)$ passing through the point of  inersection of circle $(i)$ and poin $(1,2)$ also as show in the figure.

Hence equation $(ii)$ i.e

${x^2} + {y^2} + 2x + 2y – 11 = 0\,\,\,$ is the equation of required circle.