Let a, b, c ∈ ^ + and the system of equations ( 1 - a ) x + y + z =...

We have,

$\left(1 - a\right)x + y + z = 0$

$x + \left(1 - b\right)y + z = 0$

$x + y + \left(1 - c\right)z = 0$

has many solutions~

∴ \left|\begin{cases}1 - a & 1 & 1 \\ 1 & 1 - b & 1 \\ 1 & 1 & 1 - c\end{cases}\right| = 0

$⇒ ab + ac + bc = abc$

$∴ \dfrac{ab + bc + ac}{3} \ge \left(a^{2}b^{2}c^{2}\right)1 / 3 ⇒ \left(abc\right)1 / 3 \ge 3$

$∴$~Minimum of $abc = \left(3\right)3 = 27$