The power of (1, 1) with respect to the circle $x^2+y^2-4x+3y+k=0$is 3, then k

The power of a point $P\left(x_1,y_1\right)$ with respect to the circle $S=0$ is ${\mathit{S}}_{\mathbf{11}}$

The given circle is $S\equiv x^2+y^2-4x+3y+k=0$ 

Given that The power of (1, 1) with respect to the circle $x^2+y^2-4x+3y+k=0$is 3

Hence, 

    $\begin{array}{rcl}{S}_{11}& =& 3\\ & \Rightarrow & 1+1-4+3+k=3\\ & \Rightarrow & 1+k=3\\ & \Rightarrow & k=2\end{array}$

Therefore, the value of $k$ is $\boxed{\mathbf{2}}$