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The given system of linear equations is:

\[
\begin{aligned}
1. \ 4x + 6y - 6z &= -5 \\
2. \ -4x + 9y + 2z &= 7 \\
3. \ -4x - 36y + hz &= k
\end{aligned}
\]

We are asked to determine the values of \( k \) and \( h \) such that the system has infinitely many solutions.

### Steps to solve:

1. **Condition for infinitely many solutions:**
   A system of equations will have infinitely many solutions if the equations are dependent. This occurs when one equation can be expressed as a linear combination of the others.

2. **Row Reduction Approach:**
   We will manipulate these equations, aiming to compare them and look for conditions on \( h \) and \( k \).

First, we notice that the first and third equations are similar in form. Let's reduce the system step by step:

#### Step 1: Analyze the first and second equations.
Multiply the first equation by \( -1 \):

\[
-4x - 6y + 6z = 5
\]

Compare this with the second equation:

\[
-4x + 9y + 2z = 7
\]

Subtract the two equations:

\[
(-4x - 6y + 6z) - (-4x + 9y + 2z) = 5 - 7
\]
\[
-15y + 4z = -2 \quad \text{or} \quad 15y - 4z = 2
\]

This is our new equation: \( 15y - 4z = 2 \).

#### Step 2: Analyze the first and third equations.
Multiply the first equation by \( -1 \) again:

\[
-4x - 6y + 6z = 5
\]

Compare this with the third equation:

\[
-4x - 36y + hz = k
\]

Subtract the two equations:

\[
(-4x - 6y + 6z) - (-4x - 36y + hz) = 5 - k
\]
\[
30y + (6 - h)z = 5 - k
\]

This equation becomes:

\[
30y + (6 - h)z = 5 - k
\]

#### Step 3: Set conditions for consistency.
For the system to have infinitely many solutions, the new equations must be scalar multiples of each other. Thus, the coefficients of \(

If the linear system has infinitely many solutions, then k = __ and h = __.

This problem involves solving a system of linear equations where the parameters k and h are determined so that the system has infinitely many solutions. The solution utilizes concepts of linear algebra, including row reduction and linear dependence. Suitable for college-level students.

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