Let's break down the problem step by step: ### **Step 1: Identify the normal vector** The given plane is: \[ 4y - 3x - 5z = -1 \] We can rewrite it as: \[ -3x + 4y - 5z = -1 \] The normal vector to this plane is the vector of coefficients of \( x, y, z \): \[ \vec{n} = (-3, 4, -5) \] --- ### **Step 2: Equation of the required plane** The equation of a plane passing through point \( P = (3, 4, 2) \) and having normal vector \( (-3, 4, -5) \) is: \[ -3(x - 3) + 4(y - 4) - 5(z - 2) = 0 \] --- ### **Step 3: Simplify the equation** Expand: \[ -3(x - 3) + 4(y - 4) - 5(z - 2) = 0 \\ -3x + 9 + 4y - 16 - 5z + 10 = 0 \\ -3x + 4y - 5z + 9 - 16 + 10 = 0 \\ -3x + 4y - 5z + 3 = 0 \] --- ### **Step 4: Solve for \( z \)** \[ -3x + 4y - 5z + 3 = 0 \\ -5z = 3x - 4y - 3 \\ z = \frac{3x - 4y - 3}{-5} \] Or, equivalently: \[ z = -\frac{3x}{5} + \frac{4y}{5} + \frac{3}{5} \] --- ## **Final Answer** \[ \boxed{z = -\frac{3x}{5} + \frac{4y}{5} + \frac{3}{5}} \] Let me know if you need further clarification!

Certainly! Let's go through each step in detail, including all calculations and explanations. --- ### **Problem Restatement:** Find the equation of a plane that: - Passes through the point \( P = (3, 4, 2) \), - Is parallel to the plane \( 4y - 3x - 5z = -1 \). --- ### **Step 1: Understand what it means for planes to be parallel** - Two planes are **parallel** if they have the **same normal vector**. - The **normal vector** of a plane is derived from the coefficients of \( x, y, z \) in its equation. --- ### **Step 2: Find the normal vector of the given plane** Given the plane: \[ 4y - 3x - 5z = -1 \] Rewrite in standard form: \[ -3x + 4y - 5z = -1 \] The **normal vector** \( \vec{n} \) is the vector of coefficients: \[ \boxed{ \vec{n} = (-3, 4, -5) } \] --- ### **Step 3: Write the general form of the new plane** Since the new plane is parallel to the given one, it will have the **same normal vector**: \[ \boxed{ \text{Normal vector } \vec{n} = (-3, 4, -5) } \] The general form of a plane with this normal vector passing through a point \( (x_0, y_0, z_0) \) is: \[ -3(x - x_0) + 4(y - y_0) - 5(z - z_0) = 0 \] Given \( P = (3, 4, 2) \): \[ -3(x - 3) + 4(y - 4) - 5(z - 2) = 0 \] --- ### **Step 4: Expand the equation** Let's expand each term carefully: - Expand \( -3(x - 3) \): \[ -3x + 9 \] - Expand \( 4(y - 4) \): \[ 4y - 16 \] - Expand \( -5(z - 2) \): \[ -5z + 10 \] Now, sum all these: \[ (-3x + 9) + (4y - 16) + (-5z + 10) = 0 \] Combine like terms: \[ -3x + 4y - 5z + (9 - 16 + 10) = 0 \] Calculate the constant sum: \[ 9 - 16 + 10 = 3 \] So, the **equation of the plane** is: \[ -3x + 4y - 5z + 3 = 0 \] --- ### **Step 5: Write the final equation** The equation of the required plane in standard form: \[ \boxed{ -3x + 4y - 5z + 3 = 0 } \] Alternatively, you can express \( z \) explicitly: \[ -3x + 4y - 5z + 3 = 0 \] \[ -5z = 3x - 4y - 3 \] \[ z = -\frac{3x}{5} + \frac{4y}{5} + \frac{3}{5} \] --- ### **Final answer:** \[ \boxed{ \textbf{Equation of the plane:} \quad -3x + 4y - 5z + 3 = 0 } \] or equivalently: \[ z = -\frac{3}{5}x + \frac{4}{5}y + \frac{3}{5} \] --- **Feel free to ask if you'd like more clarification on any step!**