The problem is asking for the value(s) of $h$ such that the vector $\mathbf{b} = \begin{bmatrix} 4 \\ 1 \\ h \end{bmatrix}$ lies in the plane spanned by the vectors $\mathbf{a}_1 = \begin{bmatrix} 1 \\ 4 \\ -2 \end{bmatrix}$ and $\mathbf{a}_2 = \begin{bmatrix} -2 \\ -3 \\ 7 \end{bmatrix}$.

For $\mathbf{b}$ to be in the plane spanned by $\mathbf{a}_1$ and $\mathbf{a}_2$, there must exist scalars $x_1$ and $x_2$ such that:

$x_1 \mathbf{a}_1 + x_2 \mathbf{a}_2 = \mathbf{b}$

This can be written as a system of equations by expressing the vectors explicitly:

$x_1 \begin{bmatrix} 1 \\ 4 \\ -2 \end{bmatrix} + x_2 \begin{bmatrix} -2 \\ -3 \\ 7 \end{bmatrix} = \begin{bmatrix} 4 \\ 1 \\ h \end{bmatrix}$

Breaking this into component equations, we get:

1. $x_1 - 2x_2 = 4$
2. $4x_1 - 3x_2 = 1$
3. $-2x_1 + 7x_2 = h$

We can solve the first two equations for $x_1$ and $x_2$, and then substitute those values into the third equation to solve for $h$.

Solving the system:

1. From equation 1: $x_1 = 4 + 2x_2$
2. Substitute $x_1 = 4 + 2x_2$ into equation 2:

$4(4 + 2x_2) - 3x_2 = 1$

$16 + 8x_2 - 3x_2 = 1$

$16 + 5x_2 = 1$

$5x_2 = -15$

$x_2 = -3$

3. Substitute $x_2 = -3$ into $x_1 = 4 + 2x_2$:

$x_1 = 4 + 2(-3) = 4 - 6 = -2$

Now that we have $x_1 = -2$ and $x_2 = -3$, substitute these into the third equation $-2x_1 + 7x_2 = h$:

$-2(-2) + 7(-3) = h$

$4 - 21 = h$

$h = -17$

Thus, the value of $h$ is $\boxed{-17}$.

Would you like a detailed explanation of any step or more examples? Here are 5 related questions you might find interesting:

1. What is the geometric interpretation of a vector being in the span of two other vectors?
2. How do we solve systems of linear equations using matrices?
3. Can vectors in $\mathbb{R}^3$ span more than a plane?
4. How can we check if a set of vectors are linearly independent?
5. What does it mean if the system of equations has no solution?

Tip: When dealing with systems of linear equations, using matrix methods like Gaussian elimination can help solve them efficiently.