Suppose the plane  x+ay+bz=c
  contains the point  (1,2,3)
  and the line  ℓ:r(t)=(2,6,5)+t(1,1,1)
 ,  t∈R
 . Find  a,b
  and  c
 

Given the plane equation \( x + ay + bz = c \), we need to find the constants \( a \), \( b \), and \( c \) such that the plane contains both the point \( (1, 2, 3) \) and the line \( \ell: \mathbf{r}(t) = (2, 6, 5) + t(1, 1, 1) \), where \( t \in \mathbb{R} \).

### Step 1: Use the point \( (1, 2, 3) \) on the plane
Since the point \( (1, 2, 3) \) lies on the plane, it must satisfy the plane equation:
\[
1 + 2a + 3b = c \quad \text{(Equation 1)}
\]

### Step 2: The direction vector of the line
The direction vector of the line \( \ell \) is \( \mathbf{d} = (1, 1, 1) \). Since the plane contains the line, the normal vector to the plane \( \mathbf{n} = (1, a, b) \) must be perpendicular to the direction vector \( \mathbf{d} \). Therefore, their dot product must be zero:
\[
\mathbf{n} \cdot \mathbf{d} = 1 \cdot 1 + a \cdot 1 + b \cdot 1 = 1 + a + b = 0 \quad \text{(Equation 2)}
\]

### Step 3: Use a point on the line \( \ell \)
Since the line lies entirely on the plane, any point on the line should satisfy the plane equation. For example, using the point \( (2, 6, 5) \) (which corresponds to \( t = 0 \)):
\[
2 + 6a + 5b = c \quad \text{(Equation 3)}
\]

### Step 4: Solve the system of equations
We now have the following system of equations:

1. \( 1 + 2a + 3b = c \)
2. \( 1 + a + b = 0 \)
3. \( 2 + 6a + 5b = c \)

From Equation 2:
\[
a + b = -1 \quad \Rightarrow \quad b = -1 - a
\]

Substitute \( b = -1 - a \) into Equation 1:
\[
1 + 2a + 3(-1 - a) = c
\]
\[
1 + 2a - 3 - 3a = c
\]
\[
-1 - a = c \quad \text{(Equation 4)}
\]

Substitute \( b = -1 - a \) into Equation 3:
\[
2 + 6a + 5(-1 - a) = c
\]
\[
2 + 6a - 5 - 5a = c
\]
\[
-3 + a = c \quad \text{(Equation 5)}
\]

### Step 5: Equate Equations 4 and 5
Since both equations represent \( c \):
\[
-1 - a = -3 + a
\]
\[
2a = -2 \quad \Rightarrow \quad a = -1
\]

### Step 6: Find \( b \) and \( c \)
Substitute \( a = -1 \) into \( b = -1 - a \):
\[
b = -1 - (-1) = 0
\]

Substitute \( a = -1 \) into Equation 4 or 5 to find \( c \):
\[
c = -1 - (-1) = -1
\]

### Final Answer
The values of \( a \), \( b \), and \( c \) are:
\[
a = -1, \quad b = 0, \quad c = -1
\]

Would you like more details on any part of this solution or have any questions?

Here are some related questions you might find useful:
1. How do you find the equation of a plane given three points?
2. What is the geometric interpretation of the dot product being zero?
3. How can you determine if a point lies on a given line?
4. How do you find the intersection of a line and a plane?
5. What are some applications of planes in 3D geometry?

**Tip:** When checking if a point lies on a plane, always substitute the coordinates into the plane equation to verify.

Learn how to find constants a, b, and c for the plane equation x + ay + bz = c using given conditions involving a point and a line. This problem covers concepts from linear algebra, including equations of planes and vector geometry. Suitable for Grades 10-12 students.

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