In a truncated tetrahedron the sum of no.of hexagonal faces and no.of triangular faces is

Please help Sabu decode the jail lock. Chacha Choudhary gave Sabu a formula : f_(1) = ((x)/(z)xx y), f_(2) =((f)/(v) xxu), f_(3) = ((r)/(s)xx w) Sabu can open the lock if he finds the value of 3f_(1) +f_(2) +f_(3) = key where: Number of triangular faces in a truncated tetrahedron = x Number of hexagonal faces in a truncated tetrahedron = x Number of corners in a truncated tetrahedron = z Number of square faces in a truncated octahedron = t Number of hexagonal faces in a truncated octahedron = u Number of corners in a truncated octahedron = u Number of triangular faces in a truchcated cube = w Number of octangonal faces in a truncated cube = r Number of corners in a truncated cube = s What is the KEY ?

A tetrahedron has 3 triangular faces and 1 rectangular face.

The number of hexagonal faces that are present in a truncated octahedron is

Find the number of hexagonal faces that are present in a truncated octahedral.

find the no.of triangles in given figure:

Total no.of functions