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If the four points with position vectors $-2\hat i+\hat j + \hat k, \hat i+\hat j+\hat k,\hat j- \hat k$ and $\lambda \hat j + \hat k$ are coplanar, then $\lambda$ is equal to
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If four points $\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right),\left(x_{3}, y_{3}, z_{3}\right)$ and
$\left(x_{4}, y_{4}, z_{4}\right)$ are coplanar, then $\left|\begin{array}{lll}x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \\ x_{4}-x_{1} & y_{4}-y_{1} & z_{4}-z_{1}\end{array}\right|=0$
Now, $\quad\left|\begin{array}{ccc}3 & 0 & 0 \\ 2 & 0 & -2 \\ 2 & \lambda-1 & 0\end{array}\right|=0$
$\Rightarrow \quad 3(0+2 \lambda-2)=0$
$\lambda=1$
$\left(x_{4}, y_{4}, z_{4}\right)$ are coplanar, then $\left|\begin{array}{lll}x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \\ x_{4}-x_{1} & y_{4}-y_{1} & z_{4}-z_{1}\end{array}\right|=0$
Now, $\quad\left|\begin{array}{ccc}3 & 0 & 0 \\ 2 & 0 & -2 \\ 2 & \lambda-1 & 0\end{array}\right|=0$
$\Rightarrow \quad 3(0+2 \lambda-2)=0$
$\lambda=1$
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