Step 1: Given condition for coplanarity

For three vectors to be coplanar, their scalar triple product must be zero. We have the vectors A, B, and C, and we know the given relation between λ and μ:

$$A = \lambda \hat{i} - \hat{j} + \hat{k}$$

$$B = \hat{i} + 2 \hat{j} + \mu \hat{k}$$

$$C = 3 \hat{i} - 4 \hat{j} + 5 \hat{k}$$

Scalar triple product condition:

$$[A, B, C] = A \cdot (B \times C) = 0$$

Step 2: Calculate the cross product of B and C

$$B \times C = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & \mu \\ 3 & -4 & 5 \\ \end{vmatrix}$$

$$B \times C = (10 + 4 \mu) \hat{i} - (5 - 3 \mu) \hat{j} - 10 \hat{k}$$

Step 3: Calculate the scalar triple product and apply the coplanarity condition

$$[A, B, C] = A \cdot (B \times C) = \lambda(10 + 4 \mu) - 1(5 - 3 \mu) + 1(-10) = 0$$

Step 4: Substitute the given relation between λ and μ Given that:

$$\lambda - \mu = 5$$

From the coplanarity condition, we have:

$$4 \lambda \mu + 10 \lambda - 3 \mu = 5$$

Now, substitute λ in terms of μ:

$$4(5 + \mu) \mu + 10(5 + \mu) - 3 \mu = 5$$

Step 5: Solve for μ and λ

Simplify the equation and solve for μ:

$$4 \mu^2 + 27 \mu + 45 = 0$$

Factor the equation:

$$(4 \mu + 15)(\mu + 3) = 0$$

So, the two possible values for μ are:

$$\mu = -3, \frac{-15}{4}$$

For each value of μ, find the corresponding value of λ using the given relation:

$$\lambda = \mu + 5$$

So, we get the values for λ:

$$\lambda = 2, \frac{5}{4}$$

Step 6: Calculate the sum

Now, we need to find the sum:

$$\sum\limits_{(\lambda, \mu) \in S} 80\left(\lambda^2 + \mu^2\right)$$

Substitute the values of λ and μ, and simplify:

$$80\left[(2^2 + (-3)^2) + \left(\frac{5}{4}\right)^2 + \left(\frac{-15}{4}\right)^2\right] = 80\left[13 + \frac{25}{16} + \frac{225}{16}\right]$$

$$= 80\left[13 + \frac{250}{16}\right] = 10 \times 229$$

$$= 2290$$

So, the correct answer is 2290 (Option D).

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