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, we need to follow a series of steps to find the values of \(a\), \(b\), and \(c\), and then compute \(11(a+b+c)\). Let's break it down step by step.

### Step 1: Parameterize the First Line
The first line is given by the equation:
\[ \frac{x-a}{2} = \frac{y-b}{3} = \frac{z-c}{4} \]
Let’s set this equal to a parameter \(t\):
\[ x = 2t + a, \quad y = 3t + b, \quad z = 4t + c \]

### Step 2: Parameterize the Second Line
The second line is given by:
\[ \frac{x-1}{2} = \frac{y+3}{5} = \frac{z+1}{2} \]
Let’s set this equal to a parameter \(\lambda\):
\[ x = 2\lambda + 1, \quad y = 5\lambda - 3, \quad z = 2\lambda - 1 \]

### Step 3: Find the Intersection with the Plane
The plane is given by the equation:
\[ x - y + z = 4 \]
Substituting the parameterization of the second line into the plane equation:
\[ (2\lambda + 1) - (5\lambda - 3) + (2\lambda - 1) = 4 \]
Simplifying this:
\[ 2\lambda + 1 - 5\lambda + 3 + 2\lambda - 1 = 4 \]
\[ -1\lambda + 3 = 4 \]
\[ -1\lambda = 1 \implies \lambda = -1 \]

### Step 4: Calculate the Intersection Point
Substituting \(\lambda = -1\) back into the parameterization of the second line:
\[ x = 2(-1) + 1 = -1, \quad y = 5(-1) - 3 = -8, \quad z = 2(-1) - 1 = -3 \]
Thus, the intersection point is:
\[ (-1, -8, -3) \]

### Step 5: Substitute into the Plane Equation
Now we substitute the intersection point into the plane equation \(ax - 2y + bz = 3\):
\[ a(-1) - 2(-8) + b(-3) = 3 \]
This simplifies to:
\[ -a + 16 - 3b = 3 \]
Rearranging gives:
\[ -a - 3b = -13 \implies a + 3b = 13 \quad \text{(Equation 1)} \]

### Step 6: Substitute into the First Line
Now we substitute the intersection point into the first line:
\[ \frac{-1 - a}{2} = \frac{-8 - b}{3} = \frac{-3 - c}{4} \]
Let this be equal to a parameter \(u\):
1. From \(\frac{-1 - a}{2} = u\):
\[ -1 - a = 2u \implies a = -1 - 2u \quad \text{(Equation 2)} \]
2. From \(\frac{-8 - b}{3} = u\):
\[ -8 - b = 3u \implies b = -8 - 3u \quad \text{(Equation 3)} \]
3. From \(\frac{-3 - c}{4} = u\):
\[ -3 - c = 4u \implies c = -3 - 4u \quad \text{(Equation 4)} \]

### Step 7: Substitute Equations into Equation 1
Now substitute Equations 2 and 3 into Equation 1:
\[ (-1 - 2u) + 3(-8 - 3u) = 13 \]
Expanding this:
\[ -1 - 2u - 24 - 9u = 13 \]
Combining like terms:
\[ -25 - 11u = 13 \implies -11u = 38 \implies u = -\frac{38}{11} \]

### Step 8: Calculate \(a\), \(b\), and \(c\)
Now substitute \(u\) back into Equations 2, 3, and 4:
1. For \(a\):
\[ a = -1 - 2\left(-\frac{38}{11}\right) = -1 + \frac{76}{11} = \frac{65}{11} \]
2. For \(b\):
\[ b = -8 - 3\left(-\frac{38}{11}\right) = -8 + \frac{114}{11} = \frac{26}{11} \]
3. For \(c\):
\[ c = -3 - 4\left(-\frac{38}{11}\right) = -3 + \frac{152}{11} = \frac{119}{11} \]

### Step 9: Calculate \(11(a + b + c)\)
Now we calculate:
\[ a + b + c = \frac{65}{11} + \frac{26}{11} + \frac{119}{11} = \frac{210}{11} \]
Thus,
\[ 11(a + b + c) = 11 \cdot \frac{210}{11} = 210 \]

### Final Answer
The final answer is:
\[ \boxed{210} \]