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 `p.C_21+q.C_22+r.C_23` is :

Options

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

Answer

To solve the problem, we need to find the value of \( p \cdot C_{21} + q \cdot C_{22} + r \cdot C_{23} \) where \( C_{ij} \) denotes the cofactor 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-by-Step Solution:

1. **Calculate \( C_{21} \)**:
- The cofactor \( C_{21} \) is given by:
\[ C_{21} = (-1)^{2+1} M_{21} \]
- Here, \( M_{21} \) is the minor of the element in the 2nd row and 1st column, which is obtained by removing the 2nd row and 1st column:
\[ M_{21} = \begin{vmatrix} q & r \\ m & n \end{vmatrix} = qn - rm \]
- Therefore,
\[ C_{21} = - (qn - rm) = -qn + rm \]

2. **Calculate \( C_{22} \)**:
- The cofactor \( C_{22} \) is given by:
\[ C_{22} = (-1)^{2+2} M_{22} \]
- Here, \( M_{22} \) is the minor of the element in the 2nd row and 2nd column:
\[ M_{22} = \begin{vmatrix} p & r \\ l & n \end{vmatrix} = pn - lr \]
- Therefore,
\[ C_{22} = pn - lr \]

3. **Calculate \( C_{23} \)**:
- The cofactor \( C_{23} \) is given by:
\[ C_{23} = (-1)^{2+3} M_{23} \]
- Here, \( M_{23} \) is the minor of the element in the 2nd row and 3rd column:
\[ M_{23} = \begin{vmatrix} p & q \\ l & m \end{vmatrix} = pm - ql \]
- Therefore,
\[ C_{23} = - (pm - ql) = -pm + ql \]

4. **Substituting the cofactors into the expression**:
- Now we substitute \( C_{21}, C_{22}, \) and \( C_{23} \) into the expression:
\[ p \cdot C_{21} + q \cdot C_{22} + r \cdot C_{23} \]
- This becomes:
\[ p(-qn + rm) + q(pn - lr) + r(-pm + ql) \]
- Expanding this:
\[ -pq n + prm + qp n - qlr - rpm + rql \]

5. **Combining like terms**:
- Notice that:
- The terms \( -pq n \) and \( qp n \) cancel out.
- The terms \( prm \) and \( -rpm \) cancel out.
- The terms \( -qlr \) and \( rql \) also cancel out.
- Therefore, we have:
\[ 0 \]

### Final Answer:
The value of \( p \cdot C_{21} + q \cdot C_{22} + r \cdot C_{23} \) is \( 0 \).