Let the position vectors of A,B,C,D be 5 i+5 j+2 k, i+2 j+3 k, -2 i+ j+4 k, - i+5 j+6 k.

Question

Let the position vectors of $A,B,C,D$ be $5\hat i+5\hat j+2\lambda\hat k$, $\hat i+2\hat j+3\hat k$, $-2\hat i+\lambda\hat j+4\hat k$, $-\hat i+5\hat j+6\hat k$. Let $S=\{\lambda\in\mathbb R: A,B,C,D	ext{ coplanar}\}$. Then $\sum_{\lambda\in S}(\lambda+2)^2$ is:

Accepted Answer

$\vec{AB}=(-4,-3,3-2\lambda)$, $\vec{AC}=(-7,\lambda-5,4-2\lambda)$, $\vec{AD}=(-6,0,6-2\lambda)$.

Setting $[\vec{AB}\ \vec{AC}\ \vec{AD}]=0$ and expanding leads to $\lambda^2-5\lambda+6=0 \Rightarrow \lambda=2,3$.

$$\sum(\lambda+2)^2 = 4^2+5^2 = 16+25=41$$

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