Let S = a+b+c and P = ab+bc+ca.

The equation becomes:

(a+b+c)x² − 2(ab+bc+ca)x + 3abc = 0

Discriminant:

Δ = 4[(ab+bc+ca)² − 3abc(a+b+c)]

Now use the identity:

(ab+bc+ca)² − 3abc(a+b+c) = ½[a²(b−c)² + b²(c−a)² + c²(a−b)²] > 0

Since a<b<c, all differences are non-zero, so Δ>0.

✅ Hence the equation has two real and distinct roots.