Let KK be a field and, a=((a_1),(a_2),(a_3)),b=((b_1),(b_2),(b_3)),c=((c_1),(c_2),(c_3)),d=((d_1),(d_2),(d_3)) in bbbK^3 Show that a,b,c,d are in an affine plane if and only if det((a_1,b_1,c_1,d_1),(a_2,b_2,c_2,d_3),(a_3,b_3,c_3,d_3),(1,1,1,1))=0 How can I show this?

Show that vectors are in an affine plane if and only if det=0
Let $\mathbb{K}$ be a field and, $a=\left(\begin{array}{l}{a}_{1}\\ {a}_{2}\\ {a}_{3}\end{array}\right),b=\left(\begin{array}{l}{b}_{1}\\ {b}_{2}\\ {b}_{3}\end{array}\right),c=\left(\begin{array}{c}{c}_{1}\\ {c}_{2}\\ {c}_{3}\end{array}\right),d=\left(\begin{array}{c}{d}_{1}\\ {d}_{2}\\ {d}_{3}\end{array}\right)\in {\mathbb{K}}^{3}$
Show that a,b,c,d are in an affine plane if and only if
$det\left(\begin{array}{llll}{a}_{1}& {b}_{1}& {c}_{1}& {d}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}& {d}_{2}\\ {a}_{3}& {b}_{3}& {c}_{3}& {d}_{3}\\ 1& 1& 1& 1\end{array}\right)=0$
How can I show this?
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eri1ti0m
The determinent is $det\left(\begin{array}{ccc}{b}_{1}-{a}_{1}& & \\ {b}_{2}-{a}_{2}& \cdots & \cdots \\ {b}_{3}-{a}_{3}& & \end{array}\right)=0$. Then the linear subspace $U=\mathrm{span}\left(b-a,c-a,d-a\right)$ has rank 0,1 or 2.
I think affine plane is defined to be $p+V=\left\{p+v\mid v\in V\right\}$, where the point p is in affine space and V is a linear vector (sub)space. So if $\mathrm{rank}\left(U\right)=2$, then a+U is a two dimension affine plane and $a,b,c,d\in a+U$. If $\mathrm{rank}\left(U\right)\le 1$, a+U is a affine line or point, which can contain in a plane that pass the origin (0,0,0).