If the system of linear equations
`x-2y+kz=1`
`2x+y+z=2`
`3x-y-kz=3`
has a solution `(x,y,z),z ne= 0`, then `(x.y)` lies on the straight line whose equation is :

Options

A. `3x-4y-1=0`
B. `4x-3y-4=0`
C. `4x-3y-1=0`
D. `3x-4y-4=0`

Answer

To solve the given system of linear equations and find the equation of the straight line on which the points \((x, y)\) lie, we will follow these steps:

### Step 1: Write down the equations
The system of equations is:
1. \( x - 2y + kz = 1 \) (Equation 1)
2. \( 2x + y + z = 2 \) (Equation 2)
3. \( 3x - y - kz = 3 \) (Equation 3)

### Step 2: Eliminate \(z\)
To eliminate \(z\), we can manipulate Equations 1 and 3. Let's add Equation 1 and Equation 3.

\[ (x - 2y + kz) + (3x - y - kz) = 1 + 3 \]

This simplifies to:

\[ 4x - 3y = 4 \]

### Step 3: Rearranging the equation
Now, we can rearrange this equation to express \(y\) in terms of \(x\):

\[ 4x - 3y - 4 = 0 \]

### Step 4: Find the equation of the line
From the rearranged equation, we can express it in the slope-intercept form:

\[ 3y = 4x - 4 \]
\[ y = \frac{4}{3}x - \frac{4}{3} \]

This represents a straight line in the \(xy\)-plane.

### Step 5: Conclusion
The equation of the straight line on which the points \((x, y)\) lie is:

\[ 4x - 3y - 4 = 0 \]