The equation of the plane containing the straight line ` x/2= y/3 =z/4` and perpendicular to the plane containing the straight lines `x/3=y/4 =z/2 and x/4 =y/2 =z/3` is
`px+ qy +r =0,` then `p +q +r` is

Answer

To solve the problem, we need to find the equation of a plane that contains the line given by the equation \( \frac{x}{2} = \frac{y}{3} = \frac{z}{4} \) and is perpendicular to the plane containing the lines given by \( \frac{x}{3} = \frac{y}{4} = \frac{z}{2} \) and \( \frac{x}{4} = \frac{y}{2} = \frac{z}{3} \).

### Step 1: Identify Direction Ratios of the Lines

For the line \( \frac{x}{2} = \frac{y}{3} = \frac{z}{4} \):
- The direction ratios (DR) are \( (2, 3, 4) \).

For the line \( \frac{x}{3} = \frac{y}{4} = \frac{z}{2} \):
- The direction ratios are \( (3, 4, 2) \).

For the line \( \frac{x}{4} = \frac{y}{2} = \frac{z}{3} \):
- The direction ratios are \( (4, 2, 3) \).

### Step 2: Find the Normal Vector of the Plane Containing the Two Lines

To find the normal vector of the plane containing the two lines, we can take the cross product of the direction ratios of the two lines.

Let \( \mathbf{A} = (3, 4, 2) \) and \( \mathbf{B} = (4, 2, 3) \).

The cross product \( \mathbf{A} \times \mathbf{B} \) is calculated as follows:

\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 4 & 2 \\ 4 & 2 & 3 \end{vmatrix} \]

Calculating the determinant:

\[ = \mathbf{i} (4 \cdot 3 - 2 \cdot 2) - \mathbf{j} (3 \cdot 3 - 2 \cdot 4) + \mathbf{k} (3 \cdot 2 - 4 \cdot 4) \]

\[ = \mathbf{i} (12 - 4) - \mathbf{j} (9 - 8) + \mathbf{k} (6 - 16) \]

\[ = 8\mathbf{i} - 1\mathbf{j} - 10\mathbf{k} \]

Thus, the normal vector \( \mathbf{N_1} \) of the plane containing the lines is \( (8, -1, -10) \).

### Step 3: Find the Normal Vector of the Required Plane

Next, we need to find the normal vector of the required plane, which is perpendicular to the plane containing the lines. We will take the cross product of the direction ratios of the first line \( (2, 3, 4) \) and the normal vector \( (8, -1, -10) \).

Let \( \mathbf{C} = (2, 3, 4) \).

Calculating \( \mathbf{C} \times \mathbf{N_1} \):

\[ \mathbf{C} \times \mathbf{N_1} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 4 \\ 8 & -1 & -10 \end{vmatrix} \]

Calculating the determinant:

\[ = \mathbf{i} (3 \cdot -10 - 4 \cdot -1) - \mathbf{j} (2 \cdot -10 - 4 \cdot 8) + \mathbf{k} (2 \cdot -1 - 3 \cdot 8) \]

\[ = \mathbf{i} (-30 + 4) - \mathbf{j} (-20 - 32) + \mathbf{k} (-2 - 24) \]

\[ = -26\mathbf{i} + 52\mathbf{j} - 26\mathbf{k} \]

Thus, the normal vector \( \mathbf{N_2} \) of the required plane is \( (-26, 52, -26) \).

### Step 4: Write the Equation of the Plane

The equation of the plane can be written in the form:

\[ -26x + 52y - 26z = 0 \]

We can simplify this equation by dividing by -26:

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

This can be rewritten in the form \( px + qy + rz = 0 \) where \( p = 1, q = -2, r = 1 \).

### Step 5: Calculate \( p + q + r \)

Now, we calculate \( p + q + r \):

\[ p + q + r = 1 + (-2) + 1 = 0 \]

### Final Answer

Thus, the value of \( p + q + r \) is \( \boxed{0} \).