Conditions - Linear Combination of 3 Vectors inside a Triangle (Strang P10, 1.1.20) 
Under what restrictions on $c, d, e,$ will the combinations $c\mathbf{u} + d\mathbf{v} + e\mathbf{w}$ fill in [ie bridle/rein in] the dashed triangle? To stay in the triangle, one requirement is $c \geq 0 \; \& \; d \geq 0 \; \& \; e \geq 0.$
Answer: To fill the triangle keep $c, d, e \geq 0$, and $c + d + e = 1.$


Could someone please demystify how and why $c + d + e = 1$ ? Where did it even loom from?

Supplement to Revanth Kashyap's Answer:
How and why:

¿ If $\mathbf{u}, \mathbf{v}, \mathbf{w}$ are coplanar and $k_1\mathbf{u} + k_2\mathbf{v} + k_3\mathbf{w} = 0$, then $ k_1 + k_2 + k_3 = 0$ ?

 A: Shift the Origin to the heads of any of the three vectors, say w. The original vectors now become v-w, u-w, and $\mathbf 0$(Zero Vector).
Now, in  order that a vector should lie in the triangle thus formed, it should be of the form : 
$\lambda$(v-w)+$\mu$(u-w), with $\lambda \ge 0$, $\mu$$ \ge$$ 0$ and $\lambda+\mu\le1$
Simplifying the above expression, we get :
$\mu\mathbf{u}$+$\lambda\mathbf{v}$-$(\lambda$+$\mu)\mathbf w$
Reverting to the original Origin, the expression becomes :
$\mu\mathbf{u}$+$\lambda\mathbf{v}$-$(\lambda$+$\mu)\mathbf w$ + $\mathbf w$
i.e., $\mu\mathbf{u}$+$\lambda\mathbf{v}$+$(1-(\lambda$+$\mu))\mathbf w$
Comparing it to $c\mathbf{u} + d\mathbf{v} + e\mathbf{w}$,  we get :
$\mu$ = c,  and as $\mu \ge 0$, c $\ge 0$  
$\lambda$ = d,  and as $\lambda \ge 0$, d $\ge 0$
$1-(\lambda$+$\mu)$ = e, and as $\lambda+\mu\le1$, e $\ge 0$
From the above three facts we get, c + d + e = 1 and c $\ge 0$, d $\ge 0$, and e $\ge 0$
A: consider a vector P = cU+dV+eW as a linear combination of U,V,W
Now if P has to be in the triangle as shown,
vectors UV, VW and PU have to be coplanar
so PU can be expressed as a linear combination of UV and VW
PU = mUV+nVW
(c-1)U + dV + eW = m(V-U)+ n(W-V)
(c+m-1) U + (d+n-m) V + (e-n) W = 0
since U, V , W are non-coplanar, sum of the coefficients in the above equation is 0
and hence c+d+e = 1
A: I think it should be "Since $U$, $V$ , $W$  are  not coplanar, $c+m-1=d+n-m=e-n=0$, and hence $c+d+e = 1$"
It is because $U$, $V$ , $W$ are coplanar if and only if the equations have
a solution other than α = β = γ$ = 0$ (α$U$ + β$V$ + γ$W = 0$)
