Step 1: Simplify the identity by choosing D to coincide with A.
Since the identity \(AB\times CD + BC\times AD + CA\times BD\) holds for any fourth point D, we are free to check it in the easiest case, D = A. Writing each side as a difference of position vectors (\(XY = Y-X\)), taking \(D=A\) gives \(AD = 0\), \(CD = A-C = CA\), and \(BD = A-B = -AB\).
Step 2: Substitute these into the sum.
\[
AB\times CD + BC\times AD + CA\times BD = AB\times CA + BC\times 0 + CA\times(-AB)
\]
Step 3: Simplify using the anti-symmetry of the cross product.
\[
= AB\times CA - CA\times AB = AB\times CA + AB\times CA = 2\,(AB\times CA)
\]
Step 4: Relate this to the area of triangle ABC.
Since \(CA = -AC\), we get \(AB\times CA = -(AB\times AC)\), so the magnitude is \(2|AB\times AC|\). Since the area of \(\triangle ABC\) is \(\frac{1}{2}|AB\times AC|\), we have \(|AB\times AC| = 2(\text{Area})\), so the whole sum has magnitude \(2\times 2(\text{Area}) = 4(\text{Area})\).
Step 5: State the result.
\[
\boxed{\lambda = 4}
\]