Find all three digit natural numbers of the form $(abc)_{10}$ such that $(abc)_{10}$, $(bca)_{10}$ and $(cab)_{10}$ are in geometric progression. (Here $(abc)_{10}$ is representation in base $10$.)
We need to find all three-digit natural numbers of the form $(abc)_{10}$ such that $(abc)_{10}$, $(bca)_{10}$, and $(cab)_{10}$ are in geometric progression. Let's denote these numbers as $x$, $y$, and $z$ respectively, where:
\[ x = 100a + 10b + c \]
\[ y = 100b + 10c + a \]
\[ z = 100c + 10a + b \]

Given that $x$, $y$, and $z$ are in geometric progression, we have:
\[ y^2 = xz \]

We will consider two cases: $u = v$ and $u \neq v$.

### Case 1: \( u = v \)
If \( u = v \), then from the equations:
\[ 999a = k(10u^2 - v^2) \]
\[ 999b = kv(10v - u) \]
\[ 999c = ku(10v - u) \]

Since \( u = v \), we get:
\[ 999a = k(10u^2 - u^2) = 9ku^2 \]
\[ 999b = ku(10u - u) = 9ku^2 \]
\[ 999c = ku(10u - u) = 9ku^2 \]

This implies \( a = b = c \). Therefore, the numbers are of the form $(m, m, m)$ where \( 1 \leq m \leq 9 \).

### Case 2: \( u \neq v \)
Assume \( 37 \mid k \). From the equations:
\[ 999a = k(10u^2 - v^2) \]
\[ 999b = kv(10v - u) \]
\[ 999c = ku(10v - u) \]

We have:
\[ 999 = 27 \cdot 37 \geq x = ku^2 \geq 37v^2 \]
This implies \( u^2 \leq 27 \), i.e., \( u \leq 5 \). Similarly, \( v \leq 5 \).

From the equation:
\[ 9 \mid v(v - u) \]
Since \( 1 \leq |v - u| \leq 4 \) and \( u \neq v \), we get \( 3 \mid v \) and \( 9 \mid uv \), implying \( 3 \mid u \). Therefore, \( u = v = 3 \) is impossible since \( u \neq v \).

Thus, \( 37 \nmid k \). Given \( 1 \leq u, v \leq 31 < 37 \), we have:
\[ 37 \mid 10v - u \]

There exists a non-negative integer \( t \) such that:
\[ 10v - u = 37t \]

If \( t = 0 \), then \( u = 10v \), which implies \( b = c = 0 \), a contradiction. Therefore, \( t \neq 0 \).

Moreover, we have:
\[ 37t < 10v \leq 310 \]
yielding \( t \leq 8 \).

From the equations:
\[ 27a = k(370t^2 - 200tv + 27v^2) \]
\[ 27b = ktv \]
\[ 27c = kt(10v - 37t) \]

Since \( k \) is squarefree, we have \( 9 \mid tv \), implying \( 3 \mid v \). Similarly, \( 9 \mid 370t^2 - 200tv + 27v^2 \), i.e.,
\[ 9 \mid t(t - 2v) \]

Since \( 3 \mid v \), we get \( 3 \mid t \). Therefore, \( t \in \{3, 6\} \).

#### Subcase: \( t = 6 \)
\[ u = 10v - 222 \]
This gives \( 23 \leq v \leq 25 \). Therefore, \( v = 24 \), which implies \( 3b = 16k \), a contradiction since \( 1 \leq b \leq 9 \).

#### Subcase: \( t = 3 \)
\[ u = 10v - 111 \]
This gives \( 12 \leq v \leq 14 \). Therefore, \( v = 12 \), which implies \( (u, v) = (9, 12) \).

From the equations:
\[ 3a = 2k \]
\[ 3b = 4k \]
\[ 3c = 3k \]

Thus, \( 3 \mid k \), giving two solutions:
\[ (a, b, c) = (2, 4, 3) \]
\[ (a, b, c) = (4, 8, 6) \]

### Conclusion
The system of equations has exactly 11 solutions:
\[ (a, b, c) = (m, m, m) \text{ for } 1 \leq m \leq 9 \]
\[ (a, b, c) = (2, 4, 3) \]
\[ (a, b, c) = (4, 8, 6) \]

The final answer is \( \boxed{ (a, b, c) = (m, m, m) \text{ for } 1 \leq m \leq 9 } \) and \((a, b, c) = (2, 4, 3)\) and \((a, b, c) = (4, 8, 6)\).