As, we know that, the general equation of second degree $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$ represents a pair of straight lines if and only if $\Delta=0$.
Where $\Delta=\left|\begin{array}{lll}a & h & g \\ h & b & f \\ g & f & c\end{array}\right|=a b c+2 f g h-a f^2-b g-c h^2$
Now, $2 x^2-x y+k y^2+6 x+y+4=0$ represents a
pair of straight lines, if $\left|\begin{array}{ccc}2 & -\frac{1}{2} & 3 \\ -\frac{1}{2} & k & \frac{1}{2} \\ 3 & \frac{1}{2} & 4\end{array}\right|=0$
Here, $a=$ coefficient of $x^2=2$
$$
\begin{aligned}
b & =\text { coefficient of } y^2=k \\
2 h & =\text { coefficient of } x y=-1 \\
\Rightarrow \quad h & =-\frac{1}{2} \\
2 g & =\text { coefficient of } x=6 \\
\Rightarrow \quad g & =3 \\
2 f & =\text { coefficient of } y=1 \\
\Rightarrow \quad f & =1 / 2 \\
\quad c & =\text { constant term }=4
\end{aligned}
$$
$\begin{aligned} & \Rightarrow \quad 2\left(4 k-\frac{1}{4}\right)+\frac{1}{2}\left(-2-\frac{3}{2}\right)+3\left(-\frac{1}{4}-3 k\right)=0 \\ & \begin{array}{rl}2 & 8 k-\frac{1}{2}-\frac{7}{4}-\frac{3}{4}-9 k=0 \\ \Rightarrow \quad & -k-\frac{12}{4}=0 \\ \Rightarrow \quad k=-3\end{array} \\ & \text { Hence, } 37 k^2-92 k=37 \times(-3)^2+92(-3) \\ & \quad=333-276=57\end{aligned}$