Let vec{a}, vec{b}, vec{c} be non-coplanar ve | vedclass

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(D) Let the position vectors of the three points be $\vec{P} = \lambda \vec{a}-2 \vec{b}+\vec{c}$, $\vec{Q} = 2 \vec{a}+\lambda \vec{b}-2 \vec{c}$, an

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(D) Let the position vectors of the three points be $\vec{P} = \lambda \vec{a}-2 \vec{b}+\vec{c}$, $\vec{Q} = 2 \vec{a}+\lambda \vec{b}-2 \vec{c}$, and $\vec{R} = 4 \vec{a}+7 \vec{b}-8 \vec{c}$.Since the points are collinear, the vectors $\vec{PQ}$ and $\vec{QR}$ must be parallel, i.e.,$\vec{PQ} = k \vec{QR}$ for some scalar $k$.$\vec{PQ} = \vec{Q} - \vec{P} = (2-\lambda) \vec{a} + (\lambda+2) \vec{b} - 3 \vec{c}$.$\vec{QR} = \vec{R} - \vec{Q} = (4-2) \vec{a} + (7-\lambda) \vec{b} + (-8+2) \vec{c} = 2 \vec{a} + (7-\lambda) \vec{b} - 6 \vec{c}$.Since $\vec{PQ} = k \vec{QR}$, we have:$(2-\lambda) \vec{a} + (\lambda+2) \vec{b} - 3 \vec{c} = k [2 \vec{a} + (7-\lambda) \vec{b} - 6 \vec{c}]$.Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar, they are linearly independent. Comparing the coefficients:For $\vec{c}$: $-3 = -6k \Rightarrow k = \frac{1}{2}$.For $\vec{a}$: $2-\lambda = 2k = 2(\frac{1}{2}) = 1 \Rightarrow \lambda = 1$.For $\vec{b}$: $\lambda+2 = k(7-\lambda) = \frac{1}{2}(7-\lambda) \Rightarrow 2\lambda+4 = 7-\lambda \Rightarrow 3\lambda = 3 \Rightarrow \lambda = 1$.Both conditions are satisfied for $\lambda = 1$.

Let $\vec{a}, \vec{b}, \vec{c}$ be non-coplanar vectors. If the three points with position vectors $\lambda \vec{a}-2 \vec{b}+\vec{c}$, $2 \vec{a}+\lambda \vec{b}-2 \vec{c}$, and $4 \vec{a}+7 \vec{b}-8 \vec{c}$ are collinear, then $\lambda=$

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