Circum radius of ΔDEF is 1
To find the circumradius of ΔDEF, we can use the formula:

\(R = \frac{abc}{4A}\)

where R is the circumradius, a, b, and c are the sides of the triangle, and A is the area of the triangle.

In this case, since ΔDEF is an equilateral triangle, all sides are equal to the radius of the circles that form it.

Therefore, the sides of ΔDEF are 1, 2, and 3 units.

The area of an equilateral triangle can be calculated using the formula:

\(A = \frac{\sqrt{3}}{4}a^2\)

where a is the length of the side of the triangle.

Substituting the values, we get:

\(A = \frac{\sqrt{3}}{4} \times 1^2 = \frac{\sqrt{3}}{4}\)

Now, substituting the values of a, b, c, and A into the formula for the circumradius, we get:

\(R = \frac{1 \times 2 \times 3}{4 \times \frac{\sqrt{3}}{4}} = \frac{6}{\sqrt{3}} = 2\sqrt{3}\)

Therefore, the circumradius of ΔDEF is 2√3, not 1. Hence, the statement "Circum radius of ΔDEF is 1" is incorrect.

Circumcentre of ΔABC is 1/2 units from E
The circumcentre of a triangle is the point of intersection of the perpendicular bisectors of its sides.

Since the circles touch each other externally, the centres of the circles A, B, and C are collinear with the points of tangency D, E, and F.

Therefore, the perpendicular bisector of DE passes through the point A.

Similarly, the perpendicular bisector of EF passes through the point B, and the perpendicular bisector of FD passes through the point C.

Since the perpendicular bisectors of the sides of ΔDEF intersect at the circumcentre, we can conclude that the circumcentre of ΔDEF is the same as the intersection of the perpendicular bisectors of DE, EF, and FD, which is point B.

Therefore, the statement "Circumcentre of ΔABC is 1/2 units from E" is incorrect.

Length of DE = 8/√5
To find the length of DE, we can use the Pythagorean theorem.

Since ΔDEF is an equilateral triangle, all sides are equal.

Let x be the length of DE.

Then, using the Pythagorean theorem in ΔDEF, we have:

\(x^2 + (2 + 3)^2 = (1 + 3)^2\)

\(x^2 + 25 = 16\)

\(x^2 = -9\)

Since the length of a line segment cannot be negative, it is not possible to find the length of DE.

Therefore, the statement "Length of DE = 8/√5" is incorrect.

B is the orthocentre of ΔABC
To determine if B is the orthocentre of ΔABC, we need to check if the altitudes of the triangle intersect at point B.

The altitude of a triangle is a line segment drawn from a vertex perpendicular to the