Find the equation of plane containing the lines `(x-5)/(4)=(y+7)/(4)=(z+3)/(-5) and (x-8)/(7)=(y-4)/(1)=(z-5)/(3)`.

Answer

To find the equation of the plane containing the given lines, we can follow these steps:

### Step 1: Identify the direction ratios and points on the lines
The equations of the lines are given in symmetric form:
1. Line 1: \((x-5)/4 = (y+7)/4 = (z+3)/(-5)\)
2. Line 2: \((x-8)/7 = (y-4)/1 = (z-5)/3\)

From these equations, we can extract the following information:

- For Line 1:
- Point on the line \(P_1(5, -7, -3)\)
- Direction ratios \(A_1(4, 4, -5)\)

- For Line 2:
- Point on the line \(P_2(8, 4, 5)\)
- Direction ratios \(A_2(7, 1, 3)\)

### Step 2: Find the normal vector to the plane
To find the normal vector to the plane, we can take the cross product of the direction ratios of the two lines.

Let:
\[ A_1 = \begin{pmatrix} 4 \\ 4 \\ -5 \end{pmatrix}, \quad A_2 = \begin{pmatrix} 7 \\ 1 \\ 3 \end{pmatrix} \]

The cross product \(A_1 \times A_2\) is calculated as follows:
\[ A_1 \times A_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & -5 \\ 7 & 1 & 3 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i}(4 \cdot 3 - (-5) \cdot 1) - \hat{j}(4 \cdot 3 - (-5) \cdot 7) + \hat{k}(4 \cdot 1 - 4 \cdot 7) \]
\[ = \hat{i}(12 + 5) - \hat{j}(12 + 35) + \hat{k}(4 - 28) \]
\[ = \hat{i}(17) - \hat{j}(47) - \hat{k}(24) \]
Thus, the normal vector \(N\) to the plane is:
\[ N = (17, -47, -24) \]

### Step 3: Use a point on the plane to find the equation
We can use point \(P_1(5, -7, -3)\) to find the equation of the plane. The general equation of a plane is given by:
\[ A(x - x_0) + B(y - y_0) + C(z - z_0) = 0 \]
where \((A, B, C)\) are the components of the normal vector and \((x_0, y_0, z_0)\) is a point on the plane.

Substituting the values:
\[ 17(x - 5) - 47(y + 7) - 24(z + 3) = 0 \]

Expanding this:
\[ 17x - 85 - 47y - 329 - 24z - 72 = 0 \]
\[ 17x - 47y - 24z - 485 = 0 \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ 17x - 47y - 24z + 485 = 0 \]

### Final Equation of the Plane
Thus, the equation of the plane containing the two lines is:
\[ 17x - 47y - 24z + 485 = 0 \]

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