Find an equation of the line where the planes $$ Q $$ and $$ R $$ intersect. $$ Q: x-y+3 z=1 ; R: x+y+z=0 $$ Choose the correct answer below. A. An equation of the line is given by $$ x=\frac{1}{2}-4 t, y=-\frac{1}{2}+2 t, z=2 t $$, where $$ -\infty

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To find the equation of the line where the two planes $$ Q $$ and $$ R $$ intersect, follow these steps:

### Given:
- Plane $$ Q $$: $$ x - y + 3z = 1 $$
- Plane $$ R $$: $$ x + y + z = 0 $$

### Step 1: Solve the System of Equations
Subtract the second equation from the first to eliminate $$ x $$:

$$
(x - y + 3z) - (x + y + z) = 1 - 0
$$
$$
-2y + 2z = 1 \implies y = z - \frac{1}{2}
$$

From Plane $$ R $$:

$$
x + y + z = 0 \implies x = -y - z
$$

Substitute $$ y $$ from above:

$$
x = -(z - \frac{1}{2}) - z = -2z + \frac{1}{2}
$$

### Step 2: Parameterize the Line
Let $$ z = t $$ (where $$ t $$ is a parameter). Then:

$$
x = \frac{1}{2} - 2t
$$
$$
y = t - \frac{1}{2}
$$
$$
z = t
$$

Multiplying the parameter $$ t $$ by 2 to align with the direction vector from the cross product, the parametric equations become:

$$
x = \frac{1}{2} - 4t
$$
$$
y = -\frac{1}{2} + 2t
$$
$$
z = 2t
$$

### Conclusion
The correct parametrization of the line of intersection is:

**A. An equation of the line is given by $$ x=\frac{1}{2}-4 t, y=-\frac{1}{2}+2 t, z=2 t $$, where $$ -\inftyThe correct equation of the line where planes $$ Q $$ and $$ R $$ intersect is: $$ x=\frac{1}{2}-4t $$, $$ y=-\frac{1}{2}+2t $$, $$ z=2t $$, where $$ -\infty < t < \infty $$. **Answer:** A

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