If the lines $\mathrm{x}+\mathrm{y}+1=0,4\mathrm{x}+3\mathrm{y}+4=0$ and $\mathrm{x}+\mathrm{\alpha y}+\mathrm{\beta }=0$ where ${\mathrm{\alpha }}^2+{\mathrm{\beta }}^2=2$ are concurrent then 

The given lines are 
$\begin{array}{l}\mathrm{x}+\mathrm{y}+1=0\\ 4\mathrm{x}+3\mathrm{y}+4=0\\ \mathrm{x}+\mathrm{\alpha y}+\mathrm{\beta }=0\end{array}$
If these lines are concurrent, then  $\left|\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 4\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{\alpha }\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{\beta }\end{array}\right|=0$
To solve the equation use elementary column operations 
${\mathrm{C}}_2\to {\mathrm{C}}_2-{\mathrm{C}}_1,{\mathrm{C}}_3\to {\mathrm{C}}_3-{\mathrm{C}}_1$
$$\begin{gathered} \left|\begin{array}{ccc}1& 0& 0\\ 4& -1& 0\\ 1& \mathrm{\alpha }-1& \mathrm{\beta }-1\end{array}\right|=0 \\ \begin{array}{l}1(1-\mathrm{\beta }-0)=0\\ \mathrm{\beta }=1\end{array} \end{gathered}$$
Given  ${\mathrm{\alpha }}^2+{\mathrm{\beta }}^2=2\Rightarrow {\mathrm{\alpha }}^2=1\Rightarrow \mathrm{\alpha }=\pm 1$
Therefore $\alpha =\pm 1\text{ and }\beta =1$