Two lines `(x-1)/(2)=(y-2)/(3)=(z-3)/(4)` and `(x-4)/(5)=(y-1)/(2)=(z)/(1)` intersect at a point P. If the distance of P from the plane `2x-3y+6z=7` is `lambda` units, then the value of `49lambda` is equal to

Answer

To solve the problem, we need to find the intersection point \( P \) of the two given lines and then calculate the distance of this point from the plane \( 2x - 3y + 6z = 7 \). Finally, we will compute \( 49\lambda \) where \( \lambda \) is the distance.

### Step 1: Parametrize the lines

The first line can be expressed in parametric form:
\[ \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} = t \]
From this, we can express the coordinates of the points on the line as:
\[ x = 2t + 1, \quad y = 3t + 2, \quad z = 4t + 3 \]

The second line can be expressed similarly:
\[ \frac{x-4}{5} = \frac{y-1}{2} = \frac{z}{1} = s \]
From this, we have:
\[ x = 5s + 4, \quad y = 2s + 1, \quad z = s \]

### Step 2: Set the parametric equations equal

Since both lines intersect at point \( P \), we can equate the expressions for \( x \), \( y \), and \( z \):
1. \( 2t + 1 = 5s + 4 \)
2. \( 3t + 2 = 2s + 1 \)
3. \( 4t + 3 = s \)

### Step 3: Solve the equations

From the third equation, we can express \( s \) in terms of \( t \):
\[ s = 4t + 3 \]

Substituting \( s \) into the first two equations:

1. \( 2t + 1 = 5(4t + 3) + 4 \)
\[ 2t + 1 = 20t + 15 + 4 \implies 2t + 1 = 20t + 19 \implies 18t = -18 \implies t = -1 \]

2. Substitute \( t = -1 \) into \( s = 4t + 3 \):
\[ s = 4(-1) + 3 = -4 + 3 = -1 \]

### Step 4: Find the coordinates of point \( P \)

Using \( t = -1 \) in the parametric equations of the first line:
\[ x = 2(-1) + 1 = -2 + 1 = -1 \]
\[ y = 3(-1) + 2 = -3 + 2 = -1 \]
\[ z = 4(-1) + 3 = -4 + 3 = -1 \]
Thus, the coordinates of point \( P \) are \( P(-1, -1, -1) \).

### Step 5: Calculate the distance from point \( P \) to the plane

The formula for the distance \( d \) from a point \( (x_1, y_1, z_1) \) to a plane \( Ax + By + Cz + D = 0 \) is given by:
\[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]

For the plane \( 2x - 3y + 6z - 7 = 0 \), we have:
- \( A = 2 \)
- \( B = -3 \)
- \( C = 6 \)
- \( D = -7 \)

Substituting \( P(-1, -1, -1) \):
\[ d = \frac{|2(-1) - 3(-1) + 6(-1) - 7|}{\sqrt{2^2 + (-3)^2 + 6^2}} \]
Calculating the numerator:
\[ = | -2 + 3 - 6 - 7 | = |-12| = 12 \]
Calculating the denominator:
\[ = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]
Thus, the distance \( \lambda \) is:
\[ \lambda = \frac{12}{7} \]

### Step 6: Calculate \( 49\lambda \)

Now, we calculate:
\[ 49\lambda = 49 \times \frac{12}{7} = 7 \times 12 = 84 \]

### Final Answer
The value of \( 49\lambda \) is \( \boxed{84} \).