A line `(x-a)/(2)=(y-b)/(3)=(z-c)/(4)` intersects a plane `x-y+z=4` at a point where the line `(x-1)/(2)=(y+3)/(5)=(z+1)/(2)` meets the plane. Also, a plane `ax-2y+bz=3` meet them at the same point, them `11(a+b+c)` is equal to

Answer

To solve the problem step by step, we will find the intersection of the given line and plane, and then use the intersection point to find the values of \( a \), \( b \), and \( c \).

### Step 1: Find the intersection point of the first line and the plane.

The line is given by:
\[ \frac{x-a}{2} = \frac{y-b}{3} = \frac{z-c}{4} \]
The plane is given by:
\[ x - y + z = 4 \]

We also have another line:
\[ \frac{x-1}{2} = \frac{y+3}{5} = \frac{z+1}{2} \]

Let \( \lambda \) be the parameter for the second line. Then we can express \( x \), \( y \), and \( z \) in terms of \( \lambda \):
\[ x = 2\lambda + 1, \quad y = 5\lambda - 3, \quad z = 2\lambda - 1 \]

### Step 2: Substitute into the plane equation.

Substituting \( x \), \( y \), and \( z \) into the plane equation:
\[ (2\lambda + 1) - (5\lambda - 3) + (2\lambda - 1) = 4 \]
Simplifying:
\[ 2\lambda + 1 - 5\lambda + 3 + 2\lambda - 1 = 4 \]
\[ (2\lambda - 5\lambda + 2\lambda) + (1 + 3 - 1) = 4 \]
\[ -1\lambda + 3 = 4 \]
\[ -1\lambda = 1 \implies \lambda = -1 \]

### Step 3: Find the intersection point.

Substituting \( \lambda = -1 \) back into the equations for \( x \), \( y \), and \( z \):
\[ x = 2(-1) + 1 = -1 \]
\[ y = 5(-1) - 3 = -8 \]
\[ z = 2(-1) - 1 = -3 \]
Thus, the intersection point is:
\[ (-1, -8, -3) \]

### Step 4: Substitute the intersection point into the plane \( ax - 2y + bz = 3 \).

Now we substitute \( x = -1 \), \( y = -8 \), and \( z = -3 \) into the equation:
\[ a(-1) - 2(-8) + b(-3) = 3 \]
This simplifies to:
\[ -a + 16 - 3b = 3 \]
Rearranging gives:
\[ -a - 3b + 16 = 3 \implies -a - 3b = -13 \implies a + 3b = 13 \quad \text{(1)} \]

### Step 5: Substitute the intersection point into the first line equation.

Now we substitute the intersection point into the line equation:
\[ \frac{-1 - a}{2} = \frac{-8 - b}{3} = \frac{-3 - c}{4} \]
Let this common ratio be \( \mu \):
\[ -1 - a = 2\mu \implies a = -1 - 2\mu \quad \text{(2)} \]
\[ -8 - b = 3\mu \implies b = -8 - 3\mu \quad \text{(3)} \]
\[ -3 - c = 4\mu \implies c = -3 - 4\mu \quad \text{(4)} \]

### Step 6: Substitute equations (2), (3), and (4) into equation (1).

Substituting (2) and (3) into (1):
\[ (-1 - 2\mu) + 3(-8 - 3\mu) = 13 \]
Expanding gives:
\[ -1 - 2\mu - 24 - 9\mu = 13 \]
Combining like terms:
\[ -25 - 11\mu = 13 \]
Solving for \( \mu \):
\[ -11\mu = 38 \implies \mu = -\frac{38}{11} \]

### Step 7: Find values of \( a \), \( b \), and \( c \).

Substituting \( \mu \) back into (2), (3), and (4):
\[ a = -1 - 2\left(-\frac{38}{11}\right) = -1 + \frac{76}{11} = \frac{65}{11} \]
\[ b = -8 - 3\left(-\frac{38}{11}\right) = -8 + \frac{114}{11} = \frac{26}{11} \]
\[ c = -3 - 4\left(-\frac{38}{11}\right) = -3 + \frac{152}{11} = \frac{119}{11} \]

### Step 8: Calculate \( 11(a + b + c) \).

Now, we calculate:
\[ a + b + c = \frac{65}{11} + \frac{26}{11} + \frac{119}{11} = \frac{210}{11} \]
Thus,
\[ 11(a + b + c) = 210 \]

### Final Answer:
\[ \boxed{210} \]