Consider the determinant, `Delta=|(p,q,r),(x,y,z),(l,m,n)|` .
`M_(ij)` denotes the minor of an element in `i^(th)` row, and `j^(th)` column
`C_(ij)` denotes the cofactor of an element in `i^(th)` row and `j^(th)` column
the value of `q.M_12 - y.M_22+ m.M_32` is :

Options

A. 0
B. `-Delta`
C. `Delta`
D. `Delta^2`

Answer

To solve the problem, we need to find the value of the expression \( q \cdot M_{12} - y \cdot M_{22} + m \cdot M_{32} \), where \( M_{ij} \) denotes the minor of the element in the \( i^{th} \) row and \( j^{th} \) column of the determinant \( \Delta = \begin{vmatrix} p & q & r \\ x & y & z \\ l & m & n \end{vmatrix} \).

### Step 1: Calculate the minors \( M_{12}, M_{22}, \) and \( M_{32} \)

1. **Finding \( M_{12} \)**:
- The minor \( M_{12} \) is obtained by deleting the first row and second column of the determinant.
- The remaining elements are:
\[ \begin{vmatrix} x & z \\ l & n \end{vmatrix} = x \cdot n - z \cdot l \]
- Thus, \( M_{12} = xn - zl \).

2. **Finding \( M_{22} \)**:
- The minor \( M_{22} \) is obtained by deleting the second row and second column of the determinant.
- The remaining elements are:
\[ \begin{vmatrix} p & r \\ l & n \end{vmatrix} = p \cdot n - r \cdot l \]
- Thus, \( M_{22} = pn - lr \).

3. **Finding \( M_{32} \)**:
- The minor \( M_{32} \) is obtained by deleting the third row and second column of the determinant.
- The remaining elements are:
\[ \begin{vmatrix} p & q \\ x & z \end{vmatrix} = p \cdot z - q \cdot x \]
- Thus, \( M_{32} = pz - qx \).

### Step 2: Substitute the minors into the expression

Now we substitute \( M_{12}, M_{22}, \) and \( M_{32} \) into the expression \( q \cdot M_{12} - y \cdot M_{22} + m \cdot M_{32} \):

\[ q \cdot M_{12} - y \cdot M_{22} + m \cdot M_{32} = q(xn - zl) - y(pn - lr) + m(pz - qx) \]

### Step 3: Expand the expression

Expanding the expression gives:

\[ = qxn - qzl - ypn + ylr + mpz - mqx \]

### Step 4: Rearranging the expression

Rearranging the terms, we get:

\[ = qxn - mqx - ypn + ylr + mpz - qzl \]

### Step 5: Compare with the determinant

Now, we can compare this expression with the determinant \( \Delta = \begin{vmatrix} p & q & r \\ x & y & z \\ l & m & n \end{vmatrix} \).

The determinant can be calculated as:

\[ \Delta = p(y \cdot n - z \cdot m) - q(x \cdot n - z \cdot l) + r(x \cdot m - y \cdot l) \]

Upon careful observation, we can see that the expression we derived is actually the negative of the determinant \( \Delta \).

### Conclusion

Thus, we conclude that:

\[ q \cdot M_{12} - y \cdot M_{22} + m \cdot M_{32} = -\Delta \]

The final answer is:

\[ \text{Value} = -\Delta \]