If the lines $x=1+s,\:y=3-\lambda s,\:z=1+\lambda s\:\:and\:\:x=\large\frac{t}{2}$, $y=1+t,\:z=2-t$ where $s\:\:and\:\:t$ are parameters, are $coplanar$, then $\lambda=?$

$\begin{array}{1 1} -2 \\ -1 \\ 0 \\ \large\frac{-1}{2} \end{array}$

Toolbox:
• If two lines are coplanar, then $\left |\begin {array}{ccc} x_1-x_2 & y_1-y_2 &z_1-z_2 \\l_1 &m_1 & n_1\\l_2 &m_2 & n_2\end {array}\right|=0$
The equation of the given lines can be expressed as:
$\large\frac{x-1}{1}=\frac{y-3}{-\lambda}=\frac{z-1}{\lambda}=$$s\:\:\:and\:\:\:\large\frac{x}{1/2}=\frac{y-1}{1}=\frac{z-2}{-1}=$$t$.
Here $(l_1,m_1,n_1)=(1,-\lambda,\lambda),\:\:(l_2,m_2,n_2)=(1/2,1,-1)$ and
$(x_1,y_1,z_1)=(1,3,1) \:\:and\:\:(x_2,y_2,z_2)=(0,1,2)$
Condition for the lines to be coplanar is $\left |\begin{array}{ccc} 1 & 2 & -1\\1 & -\lambda & \lambda \\1/2 & 1 & -1\end {array}\right|=0$
$\Rightarrow\:1(\lambda-\lambda)-1(-2+1)+1/2(2\lambda-\lambda)=0$
$\Rightarrow\:\lambda=-2$
answered Dec 28, 2013