If t is a parameter, then the line of intersection of the planes `3x-6y-2z=15and2x+y-2z=5` is

Options

A. `x=3+14t,y=1+2t,z=15t`
B. `x=3+14t,y=-1+2t,z=15t`
C. `x=-3+14t,y=1+2t,z=15t`
D. none of these

Answer

To find the line of intersection of the planes given by the equations \(3x - 6y - 2z = 15\) and \(2x + y - 2z = 5\), we can follow these steps:

### Step 1: Write the equations of the planes
We have the following equations:
1. \(3x - 6y - 2z = 15\) (Equation 1)
2. \(2x + y - 2z = 5\) (Equation 2)

### Step 2: Solve for one variable in terms of the others
From Equation 1, we can express \(z\) in terms of \(x\) and \(y\):
\[ 2z = 3x - 6y - 15 \implies z = \frac{3x - 6y - 15}{2} \]

### Step 3: Substitute \(z\) into the second equation
Now, substitute the expression for \(z\) into Equation 2:
\[ 2x + y - 2\left(\frac{3x - 6y - 15}{2}\right) = 5 \]
Simplifying this gives:
\[ 2x + y - (3x - 6y - 15) = 5 \]
\[ 2x + y - 3x + 6y + 15 = 5 \]
\[ -x + 7y + 15 = 5 \]
\[ -x + 7y = -10 \implies x = 7y + 10 \quad (Equation 3) \]

### Step 4: Substitute \(x\) back to find \(z\)
Now substitute \(x\) from Equation 3 back into the expression for \(z\):
\[ z = \frac{3(7y + 10) - 6y - 15}{2} \]
Simplifying this gives:
\[ z = \frac{21y + 30 - 6y - 15}{2} = \frac{15y + 15}{2} = \frac{15}{2}(y + 1) \quad (Equation 4) \]

### Step 5: Parameterize the line of intersection
Let \(y = t\) (where \(t\) is a parameter). Then from Equation 3:
\[ x = 7t + 10 \]
And from Equation 4:
\[ z = \frac{15}{2}(t + 1) \]

### Step 6: Write the parametric equations
Thus, the parametric equations of the line of intersection are:
\[ x = 7t + 10, \quad y = t, \quad z = \frac{15}{2}(t + 1) \]

### Final Result
The line of intersection of the planes can be expressed in parametric form as:
\[ \begin{align*} x &= 7t + 10 \\ y &= t \\ z &= \frac{15}{2}(t + 1) \end{align*} \]