Let ` f (x ) = | 3 - | 2- | x- 1 |||, AA x in R ` be not differentiable at ` x _ 1 , x _ 2 , x _ 3, ….x_ n ` , then `sum _ (i= 1 ) ^( n ) x _ i^2 ` equal to :

Explanation

To solve the problem, we need to find the points where the function \( f(x) = |3 - |2 - |x - 1||| \) is not differentiable. The function is not differentiable at the points where the modulus expressions change, which are known as corner points.

### Step-by-Step Solution:

1. **Identify the innermost modulus**: Start with the innermost expression, which is \( |x - 1| \). The function \( |x - 1| \) is equal to zero when \( x - 1 = 0 \), giving us:
\[ x = 1 \]

2. **Find the next modulus**: Now consider the next modulus, \( |2 - |x - 1|| \). This expression is equal to zero when:
\[ 2 - |x - 1| = 0 \implies |x - 1| = 2 \]
This leads to two cases:
- Case 1: \( x - 1 = 2 \) gives \( x = 3 \)
- Case 2: \( x - 1 = -2 \) gives \( x = -1 \)

3. **Find the outermost modulus**: Now consider the outermost expression, \( |3 - |2 - |x - 1|| | \). This is equal to zero when:
\[ 3 - |2 - |x - 1|| = 0 \implies |2 - |x - 1|| = 3 \]
Again, this leads to two cases:
- Case 1: \( 2 - |x - 1| = 3 \) gives \( |x - 1| = -1 \) (not possible)
- Case 2: \( 2 - |x - 1| = -3 \) gives \( |x - 1| = 5 \)

This leads to:
- Case 1: \( x - 1 = 5 \) gives \( x = 6 \)
- Case 2: \( x - 1 = -5 \) gives \( x = -4 \)

4. **List all points of non-differentiability**: We have found the following points where the function is not differentiable:
- \( x_1 = 1 \)
- \( x_2 = 3 \)
- \( x_3 = -1 \)
- \( x_4 = 6 \)
- \( x_5 = -4 \)

5. **Calculate the sum of squares**: Now, we need to calculate the sum of the squares of these points:
\[ \sum_{i=1}^{5} x_i^2 = 1^2 + 3^2 + (-1)^2 + 6^2 + (-4)^2 \]
Calculating each term:
- \( 1^2 = 1 \)
- \( 3^2 = 9 \)
- \( (-1)^2 = 1 \)
- \( 6^2 = 36 \)
- \( (-4)^2 = 16 \)

Now sum these values:
\[ 1 + 9 + 1 + 36 + 16 = 63 \]

### Final Answer:
The sum of the squares of the points where the function is not differentiable is:
\[ \sum_{i=1}^{n} x_i^2 = 63 \]