Explanation
**Concept:**
Let second-degree equation be ax2+by2+2hxy+2gx+2fy+c=0
It will represent a pair of straight lines, if discriminant (Δ) of this equation equals zero (Δ=0)
**Discriminant:**
Δ=ahgamp;hamp;bamp;famp;gamp;famp;c
=abc+2fgh−af2−bg2−ch2
\begin{aligned}
Calculation:
& \mathrm{Given:~Second~degree~equation,~2x^{2}+7xy+3y^{2}+8x+14y+\lambda=0} \\
& \mathrm{Compare~with~second-degree~equation~ax^{2}+by^{2}+2hxy+2gx+2fy+c=} \\
& \mathrm{So,~a=2,~b=3,~h=\frac{7}{2},~g=4,~f=7~and~c=\lambda} \\
& \text{Given equation represents a pair of straight lines} \\
& \mathrm{So,\Delta=abc+2fgh-af^{2}-bg^{2}-ch^{2}=0} \\
& \Rightarrow2\times3\times\lambda+2\times7\times4\times\frac{7}{2}-2\times(7)^{2}-3\times(4)^{2}-\lambda\times(\frac{7}{2})^{2}=0 \\
& \Rightarrow6\lambda+196-98-48-\frac{49\lambda}{4}=0 \\
& \Rightarrow50-\frac{25\lambda}{4}=0 \\
& \Rightarrow200-25\lambda=0 \\
& \therefore\lambda=8
\end{aligned}