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To tackle the problem you provided, we start by solving the two parts separately:

### Part (a): Finding a Vector Parallel to the Line of Intersection

For two planes given by the equations:
1. \(-4x + 5y - z = 0\)
2. \(-5x - 5y + 5z = 5\)

The line of intersection between these two planes can be found using the cross product of their normal vectors, which gives us a direction vector of this line.

The normal vector of the first plane is:
\(\mathbf{n}_1 = \langle -4, 5, -1 \rangle\)

The normal vector of the second plane is:
\(\mathbf{n}_2 = \langle -5, -5, 5 \rangle\)

The cross product \(\mathbf{n}_1 \times \mathbf{n}_2\) will provide the vector parallel to the line of intersection. Let's compute it:

\[
\mathbf{n}_1 \times \mathbf{n}_2 = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-4 & 5 & -1 \\
-5 & -5 & 5 \\
\end{vmatrix}
= \mathbf{i} \begin{vmatrix} 5 & -1 \\ -5 & 5 \end{vmatrix} - \mathbf{j} \begin{vmatrix} -4 & -1 \\ -5 & 5 \end{vmatrix} + \mathbf{k} \begin{vmatrix} -4 & 5 \\ -5 & -5 \end{vmatrix}
\]

Let's calculate the determinants and the resulting vector.

### Part (b): Verifying the Point and Finding the Parametric Equation

**Step 1**: Plug the point \((-1, -1, -1)\) into each plane equation to verify it lies on both planes.

**Step 2**: If it does, use the direction vector from Part (a) and this point to write the parametric equation of the line of intersection.

Let's perform the calculations and solve these steps systematically.### Results:

**Part (a):** The vector parallel to the line of intersection of the two planes is \(\langle 20, 25, 45 \rangle\).

**Part (b):** The point \((-1, -1, -1)\) lies on both planes as substituting it into both plane equations gives zero, confirming it's on the line of intersection.

### Parametric Equation of the Line of Intersection:

Using the point \((-1, -1, -1)\) and the direction vector \(\langle 20, 25, 45 \rangle\), the parametric equation of the line can be written as:

\[
\mathbf{r}(t) = \langle -1, -1, -1 \rangle + t \langle 20, 25, 45 \rangle
\]

Thus, the parametric form is:
\[
\mathbf{r}(t) = \langle -1 + 20t, -1 + 25t, -1 + 45t \rangle
\]

Would you like further details or have any questions about the steps taken here?

1. How do the coefficients in the direction vector relate to the orientation of the planes?
2. What if the cross product resulted in the zero vector?
3. How would you find the equation of the line if the point was not provided?
4. Can you use these principles to find the intersection of three planes?
5. What modifications would be needed if the planes were not in standard form?

**Tip:** Always check the feasibility of points by substituting them into the equation forms to ensure they satisfy the plane equations, helping verify the correctness of your intersection computations.

Find a vector parallel to the line of intersection of the planes -4x + 5y - z = 0 and -5x - 5y + 5z = 5. Show that the point (-1, -1, -1) lies on both planes. Then find a vector parametric equation for the line of intersection.

Learn how to find the line of intersection of the planes -4x + 5y - z = 0 and -5x - 5y + 5z = 5. Verify the point (-1, -1, -1) lies on both planes and find the parametric equation of the line. This detailed guide includes cross-product calculations and 3D vector theory. Suitable for Grades 11-12.

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