**Focal length of the objective and eyepiece**

The focal length of an objective and eyepiece of a microscope is given as 1 cm and 2 cm, respectively. The distance between them is 12 cm.

**Viewing the least distant to distinct image**

In order to view the least distant to distinct image, the object should be placed at a specific distance from the objective lens.

To determine this distance, we can use the lens formula:

1/f = 1/v - 1/u

Where:
- f is the focal length of the lens
- v is the image distance
- u is the object distance

In this case, we can consider the objective lens as the first lens, and the eyepiece as the second lens. The image formed by the objective lens becomes the object for the eyepiece.

Let's assume the distance between the object and the objective lens is x cm.

**Calculating the position of the image formed by the objective lens**

Using the lens formula for the objective lens:

1/1 = 1/v1 - 1/x

Simplifying this equation:

1/v1 = 1 - 1/x
1/v1 = (x - 1)/x
v1 = x/(x - 1)

**Calculating the position of the image formed by the eyepiece**

Now, using the lens formula for the eyepiece:

1/2 = 1/v2 - 1/(12 - x)

Simplifying this equation:

1/v2 = 1/2 + 1/(12 - x)
1/v2 = (6 + x)/(2(12 - x))
v2 = 2(12 - x)/(6 + x)

**Calculating the position of the final image**

The final image is formed by the eyepiece and should be at infinity for distinct viewing. This means that the image distance (v2) should be equal to infinity.

Let's substitute v2 with infinity in the equation:

2(12 - x)/(6 + x) = ∞

Simplifying this equation:

2(12 - x) = ∞(6 + x)
24 - 2x = ∞

Since infinity is not a finite value, we can conclude that the equation has no solution. This means that the final image cannot be formed at infinity, and therefore, the least distant to distinct image cannot be achieved.

Thus, there is no specific position where the object should be placed in order to view the least distant to distinct image in this given scenario.