Position vector of four points A, B, C, D are $-\hat i+\hat j+\hat k, 3\hat i-2\hat j + 2\hat k, 4\hat i-λ\hat j-\hat k$ and $\hat i+\hat j+\hat k$ respectively. The value of $λ$ for which the points A, B, C, D are coplanar is
Answer & explanation
Correct answer: option 2
From given points forming three vectors.
$\vec{AB}=\vec B-\vec A=4\hat i-3\hat j+\hat k$
$\vec{BC}=\vec C-\vec B=\hat i+(2-λ)\hat j-3\hat k$
$\vec{CD}=\vec D-\vec C=-3\hat i+(λ+1)\hat j+2\hat k$
for these to be coplanar their scalar triple product is zero
$\begin{vmatrix}4&-3&1\\1&2-λ&-3\\-3&λ+1&2\end{vmatrix}=0$
on solving we get $2λ+14=0$
so $λ=-7$