Three vectors in a plane are linearly dependent? The 3 vectors $u=(1,2,3)$, $v=(2,5,7)$, $w=(1,3,5)$ are linearly independent. But they should be linearly dependent as they lie on the same plane, then where this concept goes on such type of questions?
 A: It’s true that any set of three points is coplanar, but for linear dependence, we only care about planes that pass through the origin, as those are the two-dimensional vector subspaces of $\mathbb R^3$. Those planes are the set of all linear combinations of a pair of linearly-independent vectors, whereas planes that don’t include the origin can’t be generated that way. So, in the context of linear algebra, the qualifier “through the origin” is often understood and omitted to save space.
A: Adding the first row to and subtracting the second row from the third gives
$$
\begin{align}
\det\begin{bmatrix}1&2&3\\2&5&7\\1&3&5\end{bmatrix}
&=\det\begin{bmatrix}1&2&3\\2&5&7\\0&0&1\end{bmatrix}\\[6pt]
&=\det\begin{bmatrix}1&2\\2&5\end{bmatrix}\\[18pt]
&=1
\end{align}
$$
So the three vectors are linearly independent.
A: They are not in the same plane. 
Three vectors in $\mathbb{R}^3$ are in the same plane if and only if they lie in a 2-dimensional subspace of $\mathbb{R}^3$. 
Here they are linearly independent, as you correctly said, so they cannot be inside a 2-dimensional subspace.
