Key Concepts and Formulas

• Coplanarity of Vectors: Three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if their scalar triple product (STP) is zero, i.e., $[\vec{a}, \vec{b}, \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c}) = 0$.
• Scalar Triple Product using Determinant: If $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, and $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$, then $[\vec{a}, \vec{b}, \vec{c}] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$.
• Solving Systems of Equations: A system of linear and non-linear equations can be solved by substitution or elimination to find the values of unknown variables.

Step-by-Step Solution

Step 1: Set up the condition for coplanarity. The problem states that three vectors are coplanar. The condition for coplanarity of three vectors is that their scalar triple product is zero. We are given the vectors: $\vec{u} = \lambda \hat{i} - \hat{j} + \hat{k}$ $\vec{v} = \hat{i} + 2 \hat{j} + \mu \hat{k}$ $\vec{w} = 3 \hat{i} - 4 \hat{j} + 5 \hat{k}$

The scalar triple product $[\vec{u}, \vec{v}, \vec{w}]$ can be computed as the determinant of the matrix formed by their components: $[\vec{u}, \vec{v}, \vec{w}] = \begin{vmatrix} \lambda & -1 & 1 \\ 1 & 2 & \mu \\ 3 & -4 & 5 \end{vmatrix}$ For the vectors to be coplanar, this determinant must be equal to zero: $\begin{vmatrix} \lambda & -1 & 1 \\ 1 & 2 & \mu \\ 3 & -4 & 5 \end{vmatrix} = 0$ This step translates the geometric condition of coplanarity into an algebraic equation involving $\lambda$ and $\mu$.

Step 2: Expand the determinant and simplify the equation. We expand the determinant along the first row: $\lambda \left| \begin{matrix} 2 & \mu \\ -4 & 5 \end{matrix} \right| - (-1) \left| \begin{matrix} 1 & \mu \\ 3 & 5 \end{matrix} \right| + 1 \left| \begin{matrix} 1 & 2 \\ 3 & -4 \end{matrix} \right| = 0$ $\lambda((2)(5) - (\mu)(-4)) + 1((1)(5) - (\mu)(3)) + 1((1)(-4) - (2)(3)) = 0$ $\lambda(10 + 4\mu) + (5 - 3\mu) + (-4 - 6) = 0$ $10\lambda + 4\lambda\mu + 5 - 3\mu - 10 = 0$ $4\lambda\mu + 10\lambda - 3\mu - 5 = 0 \quad (*)$ This equation represents the condition on $\lambda$ and $\mu$ for the vectors to be coplanar.

Step 3: Use the given constraint to solve for $\lambda$ and $\mu$. We are given the constraint $\lambda - \mu = 5$. From this, we can express $\lambda$ in terms of $\mu$: $\lambda = \mu + 5$ Now, substitute this expression for $\lambda$ into the equation obtained in Step 2: $4(\mu+5)\mu + 10(\mu+5) - 3\mu - 5 = 0$ Expand and simplify the equation: $4\mu^2 + 20\mu + 10\mu + 50 - 3\mu - 5 = 0$ Combine like terms: $4\mu^2 + (20 + 10 - 3)\mu + (50 - 5) = 0$ $4\mu^2 + 27\mu + 45 = 0$ This is a quadratic equation in $\mu$. We can solve this by factoring. We look for two numbers that multiply to $4 \times 45 = 180$ and add up to $27$. These numbers are $12$ and $15$. $4\mu^2 + 12\mu + 15\mu + 45 = 0$ Factor by grouping: $4\mu(\mu + 3) + 15(\mu + 3) = 0$ $(\mu + 3)(4\mu + 15) = 0$ This gives two possible values for $\mu$: $\mu_1 = -3$ or $\mu_2 = -\frac{15}{4}$.

Now, we find the corresponding values of $\lambda$ using $\lambda = \mu + 5$: For $\mu_1 = -3$: $\lambda_1 = -3 + 5 = 2$. So, one pair is $(\lambda_1, \mu_1) = (2, -3)$.

For $\mu_2 = -\frac{15}{4}$: $\lambda_2 = -\frac{15}{4} + 5 = -\frac{15}{4} + \frac{20}{4} = \frac{5}{4}$. So, the other pair is $(\lambda_2, \mu_2) = \left(\frac{5}{4}, -\frac{15}{4}\right)$.

The set $S$ of all such pairs $(\lambda, \mu)$ is $S = \left\{ (2, -3), \left(\frac{5}{4}, -\frac{15}{4}\right) \right\}$. This step combines the coplanarity condition with the given linear constraint to find all valid pairs of $(\lambda, \mu)$.

Step 4: Calculate the required sum. We need to find $\sum\limits_{(\lambda, \mu) \in S} 80\left(\lambda^2+\mu^2\right)$. First, calculate $\lambda^2+\mu^2$ for each pair in $S$:

For $(\lambda_1, \mu_1) = (2, -3)$: $\lambda_1^2 + \mu_1^2 = (2)^2 + (-3)^2 = 4 + 9 = 13$.

For $(\lambda_2, \mu_2) = \left(\frac{5}{4}, -\frac{15}{4}\right)$: $\lambda_2^2 + \mu_2^2 = \left(\frac{5}{4}\right)^2 + \left(-\frac{15}{4}\right)^2 = \frac{25}{16} + \frac{225}{16} = \frac{250}{16} = \frac{125}{8}$.

Now, sum these values: $\sum\limits_{(\lambda, \mu) \in S} (\lambda^2+\mu^2) = 13 + \frac{125}{8}$ To add these, we find a common denominator: $\frac{13 \times 8}{8} + \frac{125}{8} = \frac{104}{8} + \frac{125}{8} = \frac{229}{8}$ Finally, multiply by 80: $80 \times \left(\frac{229}{8}\right) = 10 \times 229 = 2290$ This step involves performing the final calculation as requested by the problem statement.

Common Mistakes & Tips

• Sign Errors in Determinant Expansion: Be meticulous with signs when calculating the determinant. A single sign error can change the entire result.
• Algebraic Errors: Substitution and simplification of the quadratic equation are prone to arithmetic mistakes. Double-check each step, especially when dealing with fractions.
• Forgetting Roots: Ensure all roots of the quadratic equation are found, as each root corresponds to a valid pair $(\lambda, \mu)$ that satisfies the conditions.

Summary The problem requires finding pairs of $(\lambda, \mu)$ for which three given vectors are coplanar, subject to an additional linear constraint. The coplanarity condition is established using the scalar triple product, which is set to zero, leading to an algebraic equation. This equation, combined with the constraint $\lambda - \mu = 5$, forms a system that yields two pairs of $(\lambda, \mu)$. The final step involves calculating the sum $80(\lambda^2+\mu^2)$ for each pair and adding them together.

The final answer is \boxed{2290}.