Question

If the lines $\frac{x-2}{1}=\frac{y-3}{1}=\frac{Z-4}{-K}$ and $\frac{x-1}{K}=\frac{y-4}{2}=z-51$ are coplanar for $K=\alpha$ and $K=\beta$ $(\alpha \ne \beta )$, then $\alpha +\beta +6$ is equal to

• A. $3$
• B. $2$
• C. $1$
• D. $5$

Answer 1

The correct option is A $3$

$\frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}$

and $\frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2}$ are co-planar.

$\left|\begin{array}{cc}{x}_2-x_1{y}_2-y_1{z}_2-z_1& \\ l_1{m}_1{n}_1& \\ l_2{m}_2{n}_2& \end{array}\right|=0$

Now , here we have lines

$\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}$ and

$\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}$

According to co-planarity condition,

$\therefore \left|\begin{array}{cc}1-24-35-4& \\ 11-k& \\ k21& \end{array}\right|=0$

$\Rightarrow \left|\begin{array}{cc}-111& \\ 11-k& \\ k21& \end{array}\right|=0$

$\Rightarrow -1(1+2k)-1(1+k^2)+1(2-k)=0$

$\Rightarrow -1-2k-1-k^2+2-k=0$

$\Rightarrow k^2+3k=0$

$\therefore k=0$ or $k=-3$

$\therefore \alpha =0,-3$ or $\beta =-3,0$

$\therefore \alpha +\beta +\gamma =0-3+6=3$