Let `overset(to)(a_(r))=x_(r )hat(i)+y_(r )hat(j)+z_(r )hat(k),r=1,2,3` three mutually prependicular unit vectors then the value of `|{:(x_(1),,x_(2),,x_(3)),(y_(1),,y_(2) ,,y_(3)),(z_(1),,z_(2),,z_(3)):}|` is equal to

Answer

To solve the problem, we need to find the value of the determinant given by:

\[ \Delta = \begin{vmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{vmatrix} \]

where \( \overset{\to}{a_r} = x_r \hat{i} + y_r \hat{j} + z_r \hat{k} \) for \( r = 1, 2, 3 \) represents three mutually perpendicular unit vectors.

### Step 1: Write the Determinant
We start by writing the determinant explicitly:

\[ \Delta = \begin{vmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{vmatrix} \]

### Step 2: Square the Determinant
Next, we will find \( \Delta^2 \):

\[ \Delta^2 = \Delta \cdot \Delta = \begin{vmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{vmatrix} \cdot \begin{vmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{vmatrix} \]

### Step 3: Expand the Determinant
Using the property of determinants, we can expand this determinant:

\[ \Delta^2 = \begin{vmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{vmatrix} \cdot \begin{vmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{vmatrix} \]

This results in a matrix where each element is a sum of products of the components of the vectors.

### Step 4: Identify Orthogonality
Since the vectors are mutually perpendicular unit vectors, we have:

- \( x_1^2 + y_1^2 + z_1^2 = 1 \)
- \( x_2^2 + y_2^2 + z_2^2 = 1 \)
- \( x_3^2 + y_3^2 + z_3^2 = 1 \)

And the dot products between different vectors are zero:

- \( x_1x_2 + y_1y_2 + z_1z_2 = 0 \)
- \( x_1x_3 + y_1y_3 + z_1z_3 = 0 \)
- \( x_2x_3 + y_2y_3 + z_2z_3 = 0 \)

### Step 5: Form the Resultant Matrix
Thus, the determinant simplifies to:

\[ \Delta^2 = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} = 1 \]

### Step 6: Conclusion
Taking the square root gives us:

\[ \Delta = \pm 1 \]

Thus, the value of the determinant \( |{(x_1, x_2, x_3), (y_1, y_2, y_3), (z_1, z_2, z_3)}| \) is equal to \( 1 \).

### Final Answer
\[ \text{The value of the determinant is } 1. \]

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