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 `x.C_21+y.C_22+z.C_23` is :

Options

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

Answer

To solve the problem, we need to evaluate the expression \( x \cdot C_{21} + y \cdot C_{22} + z \cdot C_{23} \), 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 1: Calculate \( C_{21} \)

The cofactor \( C_{21} \) is given by:
\[ C_{21} = (-1)^{2+1} M_{21} \]
where \( M_{21} \) is the minor of the element in the 2nd row and 1st column.

To find \( M_{21} \), we remove the 2nd row and 1st column from the determinant:
\[ M_{21} = \begin{vmatrix} q & r \\ m & n \end{vmatrix} = qn - rm \]
Thus,
\[ C_{21} = - (qn - rm) \]

### Step 2: Calculate \( C_{22} \)

The cofactor \( C_{22} \) is given by:
\[ C_{22} = (-1)^{2+2} M_{22} \]
where \( M_{22} \) is the minor of the element in the 2nd row and 2nd column.

To find \( M_{22} \), we remove the 2nd row and 2nd column from the determinant:
\[ M_{22} = \begin{vmatrix} p & r \\ l & n \end{vmatrix} = pn - lr \]
Thus,
\[ C_{22} = pn - lr \]

### Step 3: Calculate \( C_{23} \)

The cofactor \( C_{23} \) is given by:
\[ C_{23} = (-1)^{2+3} M_{23} \]
where \( M_{23} \) is the minor of the element in the 2nd row and 3rd column.

To find \( M_{23} \), we remove the 2nd row and 3rd column from the determinant:
\[ M_{23} = \begin{vmatrix} p & q \\ l & m \end{vmatrix} = pm - ql \]
Thus,
\[ C_{23} = - (pm - ql) \]

### Step 4: Substitute into the expression

Now we substitute \( C_{21}, C_{22}, \) and \( C_{23} \) into the expression:
\[ x \cdot C_{21} + y \cdot C_{22} + z \cdot C_{23} = x \cdot (- (qn - rm)) + y \cdot (pn - lr) + z \cdot (- (pm - ql)) \]
This simplifies to:
\[ = -x(qn - rm) + y(pn - lr) - z(pm - ql) \]

### Step 5: Rearranging the expression

Rearranging the terms gives:
\[ = -xqn + xrm + ypn - ylr - zpm + zql \]

### Step 6: Final Result

We can express this in terms of the determinant \( \Delta \):
\[ = \Delta \]

Thus, the value of \( x \cdot C_{21} + y \cdot C_{22} + z \cdot C_{23} \) is equal to the determinant \( \Delta \).

### Final Answer
\[ \boxed{\Delta} \]