In Exercises $$ 21-22 $$, give the equation of the line that is the intersection of the given planes. plane 21. $$ p 1: 3(x-2)+(y-1)+4 z=0 $$, and $$ p 2: 2(x-1)-2(y+3)+6(z-1)=0 $$.

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To find the equation of the line that is the intersection of the two given planes, we'll follow these steps:

1. **Write the Equations of the Planes in Standard Form**
   
   The given planes are:
   $$
   \begin{align*}
   p_1 &: 3(x - 2) + (y - 1) + 4z = 0 \
   p_2 &: 2(x - 1) - 2(y + 3) + 6(z - 1) = 0
   \end{align*}
   $$
   
   Expanding these:
   $$
   \begin{align*}
   p_1 &: 3x - 6 + y - 1 + 4z = 0 \quad \Rightarrow \quad 3x + y + 4z = 7 \
   p_2 &: 2x - 2 - 2y - 6 + 6z - 6 = 0 \quad \Rightarrow \quad 2x - 2y + 6z = 14 \quad \Rightarrow \quad x - y + 3z = 7
   \end{align*}
   $$

2. **Find the Direction Vector of the Line**

The direction vector **$$\mathbf{d}$$** of the intersection line can be found by taking the cross product of the normal vectors of the two planes.

- Normal vector of $$ p_1 $$: $$\mathbf{n}_1 = [3, 1, 4]$$
   - Normal vector of $$ p_2 $$: $$\mathbf{n}_2 = [2, -2, 6]$$

$$
   \mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 = 
   \begin{vmatrix}
   \mathbf{i} & \mathbf{j} & \mathbf{k} \
   3 & 1 & 4 \
   2 & -2 & 6 \
   \end{vmatrix}
   = \mathbf{i}(1 \cdot 6 - 4 \cdot (-2)) - \mathbf{j}(3 \cdot 6 - 4 \cdot 2) + \mathbf{k}(3 \cdot (-2) - 1 \cdot 2) 
   $$
   $$
   \mathbf{d} = \mathbf{i}(6 + 8) - \mathbf{j}(18 - 8) + \mathbf{k}(-6 - 2) = \mathbf{i}(14) - \mathbf{j}(10) + \mathbf{k}(-8)
   $$
   $$
   \mathbf{d} = [14, -10, -8] = 2[7, -5, -4]
   $$
   So, a simplified direction vector is $$\mathbf{d} = [7, -5, -4]$$.

3. **Find a Point on the Line**

To find a specific point on the line, we can set one of the variables (e.g., $$ z $$) to a convenient value and solve the system of equations.

Let's set $$ z = 2 $$:

From $$ p_2 $$:
   $$
   x - y + 3(2) = 7 \quad \Rightarrow \quad x - y = 1
   $$
   
   From $$ p_1 $$:
   $$
   3x + y + 4(2) = 7 \quad \Rightarrow \quad 3x + y = -1
   $$
   
   Solving these:
   $$
   x - y = 1 \quad \text{and} \quad 3x + y = -1
   $$
   
   Adding the equations:
   $$
   4x = 0 \quad \Rightarrow \quad x = 0
   $$
   
   Substituting back:
   $$
   0 - y = 1 \quad \Rightarrow \quad y = -1
   $$
   
   So, the point is $$ (0, -1, 2) $$.

4. **Write the Parametric Equations of the Line**

Using the point $$ (0, -1, 2) $$ and direction vector $$ [7, -5, -4] $$, the parametric equations are:
   $$
   \begin{cases}
   x = 0 + 7t \
   y = -1 - 5t \
   z = 2 - 4t
   \end{cases}
   $$
   where $$ t $$ is a parameter.

**Final Answer:**

A parametric equation for the intersection line is

x = 7 t,

y = –1 – 5 t,

and z = 2 – 4 t

for real numbers t.
A parametric equation for the intersection line is: $$ \begin{cases} x = 7t \ y = -1 - 5t \ z = 2 - 4t \end{cases} $$ where $$ t $$ is a real number.

In Exercises \( 21-22 \), give the equation of the line that is the intersection of the given planes. plane 21. \( p 1: 3(x-2)+(y-1)+4 z=0 \), and \( p 2: 2(x-1)-2(y+3)+6(z-1)=0 \).

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