You are given the following two planes:

x+4y-3z-12=0
x+6y-2z-22=0

a) Use the two original equations to determine two other equations that have the same solution as the original two.

ANS: the original two are not parallel in anyway so can I just take multiples of the original two OR is there any other way to find them?

b) Find a third equation that will have a unique solution with the original two equations.

a) To determine two other equations that have the same solution as the original two, you can take linear combinations of the original equations. A linear combination of two equations involves multiplying each equation by a constant and adding them together.

Let's take the first equation (E1):

E1: x + 4y - 3z - 12 = 0

Now, let's take the second equation (E2):

E2: x + 6y - 2z - 22 = 0

To find two equations that have the same solution, we can take linear combinations of E1 and E2 by multiplying each equation by a constant and adding them together.

For example, let's multiply E1 by 2 and E2 by 3, and then add them together:

2(E1) = 2x + 8y - 6z - 24 = 0

3(E2) = 3x + 18y - 6z - 66 = 0

Adding them together, we get:

2x + 8y - 6z - 24 + 3x + 18y - 6z - 66 = 0

Simplifying, we have:

5x + 26y - 12z - 90 = 0

So, the equation 5x + 26y - 12z - 90 = 0 is a linear combination of the original two equations and has the same solution.

You can repeat this process with other multiples or combinations of E1 and E2 to find more equations that have the same solution.

b) To find a third equation that will have a unique solution with the original two equations, you need to find the intersection point of the two planes represented by the equations.

To find the intersection point, we can solve the system of equations formed by the two given equations:

Equation 1: x + 4y - 3z - 12 = 0

Equation 2: x + 6y - 2z - 22 = 0

There are multiple methods to solve this system of equations, such as substitution, elimination, or using matrices. Let's use the elimination method here.

Multiplying Equation 1 by 3 and Equation 2 by 2, we get:

3(x + 4y - 3z - 12) = 0
2(x + 6y - 2z - 22) = 0

Simplifying, we have:

3x + 12y - 9z - 36 = 0
2x + 12y - 4z - 44 = 0

Now, let's subtract Equation 2 from Equation 1:

(3x + 12y - 9z - 36) - (2x + 12y - 4z - 44) = 0

Simplifying, we get:

x - 5z + 8 = 0

So, the third equation that will have a unique solution with the original two equations is x - 5z + 8 = 0.