On comparing the given equations with $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$, we get $a=2, b=2 \lambda, h=-5, g=\frac{5}{2}$, $f=-8, c=-3$
Since, the given equations represents a pair of straight lines, therefore
$$
\begin{aligned}
& \left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=0 \\
& \Rightarrow\left|\begin{array}{ccc}
2 & -5 & 5 / 2 \\
-5 & 2 \lambda & -8 \\
5 / 2 & -8 & -3
\end{array}\right|=0 \\
& \Rightarrow 2(-6 \lambda-64)+5(15+20)+\frac{5}{2}(40-5 \lambda)=0 \\
& \Rightarrow-12 \lambda-128+175+100-\frac{25}{2} \lambda=0 \\
& \Rightarrow \quad-\frac{49}{2} \lambda=-147 \\
& \Rightarrow \quad \lambda=\frac{147 \times 2}{49}=6 \\
&
\end{aligned}
$$
Now, the point of intersection of given lines is given by
$$
\left(\frac{b g-f h}{h^2-a b}, \frac{a f-g h}{h^2-a b}\right)=\left(\frac{30-40}{25-24}, \frac{-16+\frac{25}{2}}{25-24}\right)
$$

$$
=\left(-10, \frac{-7}{2}\right)
$$