The value of `k in R,`for which the following system of linear equation
`3x - y + 4z = 3,`
`x + 2y - 3z = - 2,`
`6x + 5y + kz = - 3,`
Has infinitely many solutions, is:

Options

A. 5
B. 3
C. `-5`
D. `-3`

Answer

To find the value of \( k \) for which the given system of linear equations has infinitely many solutions, we need to analyze the system of equations:

1. \( 3x - y + 4z = 3 \)
2. \( x + 2y - 3z = -2 \)
3. \( 6x + 5y + kz = -3 \)

For a system of linear equations to have infinitely many solutions, the determinant of the coefficients must be zero. This means we need to calculate the determinant of the coefficient matrix formed by the coefficients of \( x, y, z \) from the equations.

The coefficient matrix is:
\[ \begin{bmatrix} 3 & -1 & 4 \\ 1 & 2 & -3 \\ 6 & 5 & k \end{bmatrix} \]

The determinant of this matrix can be calculated using the formula for the determinant of a 3x3 matrix:
\[ \text{Det} = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is:
\[ \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \]

Substituting the values from our matrix:
- \( a = 3, b = -1, c = 4 \)
- \( d = 1, e = 2, f = -3 \)
- \( g = 6, h = 5, i = k \)

Now, we can calculate the determinant:
\[ \text{Det} = 3(2k - (-3)(5)) - (-1)(1 \cdot k - (-3)(6)) + 4(1 \cdot 5 - 2 \cdot 6) \]

Calculating each term:
1. First term: \( 3(2k + 15) = 6k + 45 \)
2. Second term: \( -(-1)(k + 18) = k + 18 \)
3. Third term: \( 4(5 - 12) = 4(-7) = -28 \)

Now, combining these:
\[ \text{Det} = (6k + 45) + (k + 18) - 28 \]
\[ = 6k + k + 45 + 18 - 28 \]
\[ = 7k + 35 \]

Setting the determinant equal to zero for infinitely many solutions:
\[ 7k + 35 = 0 \]
\[ 7k = -35 \]
\[ k = -5 \]

Thus, the value of \( k \) for which the system has infinitely many solutions is:
\[ \boxed{-5} \]