The equation of the circle are
$$
\begin{array}{r}
x^2+y^2-2 \lambda x-2 y-7=0 \\
3\left(x^2+y^2\right)-8 x+29 y=0 \\
x^2+y^2-\frac{8}{3} x+\frac{29}{3} y=0
\end{array}
$$
On comparing Eq. (i) and Eq. (ii) with
$$
\begin{array}{r}
x^2+y^2+2 g x+2 f y+c=0, \text { we get } \\
g_1=-\lambda, f_1=-1, c_1=-7
\end{array}
$$

and
$$
g_2=-\frac{4}{3}, f_2=\frac{29}{6}, c_2=0
$$
Since, circles are orthogonal, then
$$
\begin{array}{lc}
& 2 g_1 g_2+2 f_1 f_2=c_1+c_2 \\
\Rightarrow & 2(-\lambda)\left(\frac{-4}{3}\right)+2(-1)\left(\frac{29}{6}\right)=-7+0 \\
\Rightarrow & \frac{8}{3} \lambda-\frac{29}{3}=-7 \Rightarrow \frac{8}{3} \lambda=-7+\frac{29}{3} \\
\Rightarrow & \frac{8}{3} \lambda=\frac{-21+29}{3} \Rightarrow \lambda=\frac{8}{3} \times \frac{3}{8} \\
\therefore & \quad \lambda=1
\end{array}
$$