If `Delta_1=|{:(10,4,3),(17,7,4),(4,-5,7):}|,Delta_2=|{:(4,x+5,3),(7,x+12,4),(-5,x-1,7):}|` such that `Delta_1+Delta_2=0,` then

Options

A. `x=5`
B. x has no real value
C. `x=0`
D. `x=1`

Answer

To solve the problem, we need to evaluate the determinants \( \Delta_1 \) and \( \Delta_2 \) and use the condition that \( \Delta_1 + \Delta_2 = 0 \).

### Step 1: Calculate \( \Delta_1 \)

Given:
\[ \Delta_1 = \begin{vmatrix} 10 & 4 & 3 \\ 17 & 7 & 4 \\ 4 & -5 & 7 \end{vmatrix} \]

To find \( \Delta_1 \), we can use the formula for the determinant of a 3x3 matrix:
\[ \Delta = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is:
\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} \]

In our case:
- \( a = 10, b = 4, c = 3 \)
- \( d = 17, e = 7, f = 4 \)
- \( g = 4, h = -5, i = 7 \)

Calculating \( \Delta_1 \):
\[ \Delta_1 = 10 \begin{vmatrix} 7 & 4 \\ -5 & 7 \end{vmatrix} - 4 \begin{vmatrix} 17 & 4 \\ 4 & 7 \end{vmatrix} + 3 \begin{vmatrix} 17 & 7 \\ 4 & -5 \end{vmatrix} \]

Calculating the smaller 2x2 determinants:
1. \( \begin{vmatrix} 7 & 4 \\ -5 & 7 \end{vmatrix} = (7 \cdot 7) - (4 \cdot -5) = 49 + 20 = 69 \)
2. \( \begin{vmatrix} 17 & 4 \\ 4 & 7 \end{vmatrix} = (17 \cdot 7) - (4 \cdot 4) = 119 - 16 = 103 \)
3. \( \begin{vmatrix} 17 & 7 \\ 4 & -5 \end{vmatrix} = (17 \cdot -5) - (7 \cdot 4) = -85 - 28 = -113 \)

Now substituting back:
\[ \Delta_1 = 10(69) - 4(103) + 3(-113) \]
\[ = 690 - 412 - 339 \]
\[ = 690 - 751 = -61 \]

### Step 2: Calculate \( \Delta_2 \)

Given:
\[ \Delta_2 = \begin{vmatrix} 4 & x+5 & 3 \\ 7 & x+12 & 4 \\ -5 & x-1 & 7 \end{vmatrix} \]

Using the same determinant formula:
\[ \Delta_2 = 4 \begin{vmatrix} x+12 & 4 \\ x-1 & 7 \end{vmatrix} - (x+5) \begin{vmatrix} 7 & 4 \\ -5 & 7 \end{vmatrix} + 3 \begin{vmatrix} 7 & x+12 \\ -5 & x-1 \end{vmatrix} \]

Calculating the smaller 2x2 determinants:
1. \( \begin{vmatrix} x+12 & 4 \\ x-1 & 7 \end{vmatrix} = (x+12) \cdot 7 - 4 \cdot (x-1) = 7x + 84 - 4x + 4 = 3x + 88 \)
2. We already calculated \( \begin{vmatrix} 7 & 4 \\ -5 & 7 \end{vmatrix} = 69 \)
3. \( \begin{vmatrix} 7 & x+12 \\ -5 & x-1 \end{vmatrix} = 7(x-1) - (x+12)(-5) = 7x - 7 + 5x + 60 = 12x + 53 \)

Now substituting back:
\[ \Delta_2 = 4(3x + 88) - (x + 5)(69) + 3(12x + 53) \]
\[ = 12x + 352 - 69x - 345 + 36x + 159 \]
\[ = (12x - 69x + 36x) + (352 - 345 + 159) \]
\[ = -21x + 166 \]

### Step 3: Set up the equation \( \Delta_1 + \Delta_2 = 0 \)

From the problem statement:
\[ \Delta_1 + \Delta_2 = 0 \implies -61 + (-21x + 166) = 0 \]
\[ -21x + 105 = 0 \]
\[ 21x = 105 \]
\[ x = 5 \]

### Final Answer
The value of \( x \) is \( 5 \).