Without expanding at any stage, find the value of the determinant :
`Delta=|(20,a,b+c),(20,b,a+c),(20,c,a+b)|`

Answer

To find the value of the determinant

\[ \Delta = \begin{vmatrix} 20 & a & b+c \\ 20 & b & a+c \\ 20 & c & a+b \end{vmatrix} \]

we can use properties of determinants without expanding it.

### Step 1: Apply a Column Operation
We can change the third column (C3) by adding the second column (C2) to it. This operation does not change the value of the determinant.

\[ C3 \rightarrow C2 + C3 \]

After applying this operation, the determinant becomes:

\[ \Delta = \begin{vmatrix} 20 & a & b+c + a \\ 20 & b & a+c + b \\ 20 & c & a+b + c \end{vmatrix} \]

This simplifies to:

\[ \Delta = \begin{vmatrix} 20 & a & a+b+c \\ 20 & b & a+b+c \\ 20 & c & a+b+c \end{vmatrix} \]

### Step 2: Factor Out Common Elements
Now, we can factor out the common elements from the first column. The first column has the same element (20) in all rows.

\[ \Delta = 20 \cdot \begin{vmatrix} 1 & a & a+b+c \\ 1 & b & a+b+c \\ 1 & c & a+b+c \end{vmatrix} \]

### Step 3: Observe the Structure of the Determinant
Notice that the first and second columns are identical in the determinant:

\[ \Delta = 20 \cdot \begin{vmatrix} 1 & a & a+b+c \\ 1 & b & a+b+c \\ 1 & c & a+b+c \end{vmatrix} \]

### Step 4: Conclude the Value of the Determinant
Since two rows of the determinant are the same (the first column is all 1's), the determinant evaluates to 0.

Thus, we have:

\[ \Delta = 20 \cdot 0 = 0 \]

### Final Answer
The value of the determinant \(\Delta\) is

\[ \Delta = 0 \]

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