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 step by step, we will find the point of intersection of the two lines and then calculate the distance from that point to the given plane.

### Step 1: Parameterize the lines
The first line is given by:
\[ \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} \]
Let \( t \) be the parameter. Then we can express the coordinates as:
\[ x = 2t + 1, \quad y = 3t + 2, \quad z = 4t + 3 \]

The second line is given by:
\[ \frac{x-4}{5} = \frac{y-1}{2} = \frac{z}{1} \]
Let \( s \) be the parameter. Then we can express the coordinates as:
\[ x = 5s + 4, \quad y = 2s + 1, \quad z = s \]

### Step 2: Set the equations equal to find the intersection
Since both lines intersect at point \( P \), we can set the equations equal to each other:
\[ 2t + 1 = 5s + 4 \quad (1) \]
\[ 3t + 2 = 2s + 1 \quad (2) \]
\[ 4t + 3 = s \quad (3) \]

### Step 3: Solve the equations
From equation (3):
\[ s = 4t + 3 \]
Substituting \( s \) into equations (1) and (2):

**Substituting into equation (1):**
\[ 2t + 1 = 5(4t + 3) + 4 \]
\[ 2t + 1 = 20t + 15 + 4 \]
\[ 2t + 1 = 20t + 19 \]
\[ 2t - 20t = 19 - 1 \]
\[ -18t = 18 \implies t = -1 \]

**Substituting \( t = -1 \) into equation (3) to find \( s \):**
\[ s = 4(-1) + 3 = -4 + 3 = -1 \]

### Step 4: Find the coordinates of point \( P \)
Now substituting \( t = -1 \) into the parameterization 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 point \( P \) is \( (-1, -1, -1) \).

### Step 5: Calculate the distance from point \( P \) to the plane
The equation of the plane is:
\[ 2x - 3y + 6z = 7 \]
To find the distance \( d \) from point \( P(-1, -1, -1) \) to the plane, we use the formula:
\[ d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
where \( A = 2, B = -3, C = 6, D = -7 \) (from the plane equation rearranged to \( 2x - 3y + 6z - 7 = 0 \)).

Substituting \( x_1 = -1, y_1 = -1, z_1 = -1 \):
\[ d = \frac{|2(-1) - 3(-1) + 6(-1) - 7|}{\sqrt{2^2 + (-3)^2 + 6^2}} \]
Calculating the numerator:
\[ = | -2 + 3 - 6 - 7 | = |-12| = 12 \]
Calculating the denominator:
\[ = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]
Thus,
\[ d = \frac{12}{7} \]

### Step 6: Calculate \( 49\lambda \)
Since \( \lambda = \frac{12}{7} \), we find:
\[ 49\lambda = 49 \cdot \frac{12}{7} = 7 \cdot 12 = 84 \]

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