The perpendicular distance from the origin to the plane containing the two lines, `(x +2)/( 3) = ( y -2)/(5) = (z + 5)/(7) and (x-1)/(1) = (y-4)/(4) = (z+4)/(7),` is `(p)/(sqrtq),` then pq is

Answer

To find the perpendicular distance from the origin to the plane containing the two given lines, we will follow these steps:

### Step 1: Identify Direction Ratios and Points
The direction ratios of the first line \((x + 2)/3 = (y - 2)/5 = (z + 5)/7\) are \(3, 5, 7\).
The direction ratios of the second line \((x - 1)/1 = (y - 4)/4 = (z + 4)/7\) are \(1, 4, 7\).

The points on the lines are:
- For the first line, a point \(P_1 = (-2, 2, -5)\).
- For the second line, a point \(P_2 = (1, 4, -4)\).

### Step 2: Write the Equation of the Plane
To find the equation of the plane containing both lines, we can use the determinant method. The general form of the equation of a plane is given by:

\[ \begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0 \]

Where \((x_1, y_1, z_1)\) is a point on the plane, and \((a_1, b_1, c_1)\) and \((a_2, b_2, c_2)\) are the direction ratios of the two lines.

Substituting in our values:
- Point \(P_1 = (-2, 2, -5)\)
- Direction ratios: \((3, 5, 7)\) and \((1, 4, 7)\)

The determinant becomes:

\[ \begin{vmatrix} x + 2 & y - 2 & z + 5 \\ 3 & 5 & 7 \\ 1 & 4 & 7 \end{vmatrix} = 0 \]

### Step 3: Expand the Determinant
Expanding the determinant gives:

\[ (x + 2) \begin{vmatrix} 5 & 7 \\ 4 & 7 \end{vmatrix} - (y - 2) \begin{vmatrix} 3 & 7 \\ 1 & 7 \end{vmatrix} + (z + 5) \begin{vmatrix} 3 & 5 \\ 1 & 4 \end{vmatrix} = 0 \]

Calculating each determinant:

1. \(\begin{vmatrix} 5 & 7 \\ 4 & 7 \end{vmatrix} = 5 \cdot 7 - 4 \cdot 7 = 35 - 28 = 7\)

2. \(\begin{vmatrix} 3 & 7 \\ 1 & 7 \end{vmatrix} = 3 \cdot 7 - 1 \cdot 7 = 21 - 7 = 14\)

3. \(\begin{vmatrix} 3 & 5 \\ 1 & 4 \end{vmatrix} = 3 \cdot 4 - 1 \cdot 5 = 12 - 5 = 7\)

Substituting back into the equation gives:

\[ (x + 2) \cdot 7 - (y - 2) \cdot 14 + (z + 5) \cdot 7 = 0 \]

### Step 4: Simplify the Plane Equation
Expanding and simplifying:

\[ 7x + 14 - 14y + 28 + 7z + 35 = 0 \]

Combining like terms:

\[ 7x - 14y + 7z + 77 = 0 \]

Dividing through by 7 gives:

\[ x - 2y + z + 11 = 0 \]

### Step 5: Find the Distance from the Origin to the Plane
The formula for the distance \(D\) from a point \((x_0, y_0, z_0)\) to the plane \(Ax + By + Cz + D = 0\) is:

\[ D = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

For our plane \(x - 2y + z + 11 = 0\) (where \(A = 1\), \(B = -2\), \(C = 1\), and \(D = 11\)), and the origin \((0, 0, 0)\):

\[ D = \frac{|1(0) - 2(0) + 1(0) + 11|}{\sqrt{1^2 + (-2)^2 + 1^2}} = \frac{|11|}{\sqrt{1 + 4 + 1}} = \frac{11}{\sqrt{6}} \]

### Step 6: Find \(pq\)
Here, \(p = 11\) and \(q = 6\). Therefore, \(pq = 11 \times 6 = 66\).

### Final Answer
Thus, the value of \(pq\) is \(66\).

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