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

To find the intersection of the given planes, we can first express each plane equation in standard form. For plane \( p_1 \): \[ 3(x-2) + (y-1) + 4z = 0 \] Expanding this gives: \[ 3x - 6 + y - 1 + 4z = 0 \] or, \[ 3x + y + 4z = 7 \] For plane \( p_2 \): \[ 2(x-1) - 2(y+3) + 6(z-1) = 0 \] Expanding this gives: \[ 2x - 2 - 2y - 6 + 6z - 6 = 0 \] or, \[ 2x - 2y + 6z = 14 \] Now we have the system of equations: 1. \( 3x + y + 4z = 7 \) 2. \( 2x - 2y + 6z = 14 \) We can use substitution or elimination to find the line of intersection. For this case, let's express \( y \) from the first equation: \[ y = 7 - 3x - 4z \] Now we substitute this into the second equation: \[ 2x - 2(7 - 3x - 4z) + 6z = 14 \] Simplifying it: \[ 2x - 14 + 6x + 8z + 6z = 14 \] So we get: \[ 8x + 14z = 28 \] Dividing by 4 gives: \[ 2x + \frac{14}{4}z = 7 \] Or: \[ 2x + 3.5z = 7 \] Expressing \( x \): \[ x = \frac{7 - 3.5z}{2} \] Now substitute \( x \) back into the equation for \( y \): \[ y = 7 - 3\left(\frac{7 - 3.5z}{2}\right) - 4z \] Now let \( z = t \) (a parameter). Then the parametric equations for the line of intersection are: 1. \( x = \frac{7 - 3.5t}{2} \) 2. \( y = 7 - 3\left(\frac{7 - 3.5t}{2}\right) - 4t \) 3. \( z = t \) This gives the equation of the line of intersection in parametric form. Now you can simplify it further if needed or just use it as is for calculations! Enjoy exploring the beauty of geometry!