A solution set of the equations `x+2y+z=1`, `x+3y+4z=k`, `x+5y+10z=k^(2)` is

Options

A. `(1+5lambda,-3lambda,lambda)`
B. `(5lambda-1,1-3lambda,lambda)`
C. `(1+6lambda,-2lambda,lambda)`
D. `(1-6lambda,lambda,lambda)`

Explanation

To find the solution set of the equations:

1. \( x + 2y + z = 1 \) (Equation 1)
2. \( x + 3y + 4z = k \) (Equation 2)
3. \( x + 5y + 10z = k^2 \) (Equation 3)

we will follow these steps:

### Step 1: Subtract Equation 1 from Equation 2
We start by subtracting Equation 1 from Equation 2 to eliminate \( x \):

\[ (x + 3y + 4z) - (x + 2y + z) = k - 1 \]

This simplifies to:

\[ y + 3z = k - 1 \quad \text{(Equation 4)} \]

### Step 2: Subtract Equation 2 from Equation 3
Next, we subtract Equation 2 from Equation 3:

\[ (x + 5y + 10z) - (x + 3y + 4z) = k^2 - k \]

This simplifies to:

\[ 2y + 6z = k^2 - k \quad \text{(Equation 5)} \]

### Step 3: Simplify Equation 5
We can factor out a 2 from Equation 5:

\[ 2(y + 3z) = k^2 - k \]

Dividing both sides by 2 gives:

\[ y + 3z = \frac{k^2 - k}{2} \quad \text{(Equation 6)} \]

### Step 4: Set Equations 4 and 6 Equal
Now, we have two expressions for \( y + 3z \):

From Equation 4:
\[ y + 3z = k - 1 \]

From Equation 6:
\[ y + 3z = \frac{k^2 - k}{2} \]

Setting these equal gives:

\[ k - 1 = \frac{k^2 - k}{2} \]

### Step 5: Clear the Fraction
Multiply through by 2 to eliminate the fraction:

\[ 2(k - 1) = k^2 - k \]

This simplifies to:

\[ 2k - 2 = k^2 - k \]

Rearranging gives:

\[ k^2 - 3k + 2 = 0 \]

### Step 6: Factor the Quadratic
We can factor this quadratic equation:

\[ (k - 1)(k - 2) = 0 \]

Thus, the solutions for \( k \) are:

\[ k = 1 \quad \text{or} \quad k = 2 \]

### Step 7: Find Solutions for Each Value of \( k \)

#### Case 1: \( k = 1 \)
Substituting \( k = 1 \) into Equation 4:

\[ y + 3z = 1 - 1 \implies y + 3z = 0 \implies y = -3z \]

Substituting \( y = -3z \) into Equation 1:

\[ x + 2(-3z) + z = 1 \implies x - 6z + z = 1 \implies x - 5z = 1 \implies x = 1 + 5z \]

Thus, the solution set is:

\[ (x, y, z) = (1 + 5\lambda, -3\lambda, \lambda) \quad \text{where } \lambda = z \]

#### Case 2: \( k = 2 \)
Substituting \( k = 2 \) into Equation 4:

\[ y + 3z = 2 - 1 \implies y + 3z = 1 \implies y = 1 - 3z \]

Substituting \( y = 1 - 3z \) into Equation 1:

\[ x + 2(1 - 3z) + z = 1 \implies x + 2 - 6z + z = 1 \implies x - 5z + 2 = 1 \implies x = 5z - 1 \]

Thus, the solution set is:

\[ (x, y, z) = (5\lambda - 1, 1 - 3\lambda, \lambda) \quad \text{where } \lambda = z \]

### Final Solution Sets
The two sets of solutions are:

1. \( (1 + 5\lambda, -3\lambda, \lambda) \)
2. \( (5\lambda - 1, 1 - 3\lambda, \lambda) \)