If the vectors `vec a=lambdahat i+muhat j+4hat k,vec b=-2hat i+4hat j-2hat k,vec c=2hat i+3hat j+hat k` are coplanar and the projection of `vec a` on the vector `vec b` is `sqrt(54)` units,then the sum of all possible values of `lambda+mu` is equal to

Options

A. `18`
B. `6`
C. `24`
D. `0`

Answer

To solve the problem, we need to follow these steps:

### Step 1: Establish the condition for coplanarity
The vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are coplanar if the determinant of the matrix formed by their components is zero.

The vectors are given as:
\[ \vec{a} = \lambda \hat{i} + \mu \hat{j} + 4 \hat{k} \]
\[ \vec{b} = -2 \hat{i} + 4 \hat{j} - 2 \hat{k} \]
\[ \vec{c} = 2 \hat{i} + 3 \hat{j} + 1 \hat{k} \]

The determinant can be set up as follows:
\[ \begin{vmatrix} \lambda & \mu & 4 \\ -2 & 4 & -2 \\ 2 & 3 & 1 \end{vmatrix} = 0 \]

### Step 2: Calculate the determinant
Calculating the determinant:
\[ \lambda \begin{vmatrix} 4 & -2 \\ 3 & 1 \end{vmatrix} - \mu \begin{vmatrix} -2 & -2 \\ 2 & 1 \end{vmatrix} + 4 \begin{vmatrix} -2 & 4 \\ 2 & 3 \end{vmatrix} = 0 \]

Calculating the 2x2 determinants:
1. \(\begin{vmatrix} 4 & -2 \\ 3 & 1 \end{vmatrix} = (4)(1) - (-2)(3) = 4 + 6 = 10\)
2. \(\begin{vmatrix} -2 & -2 \\ 2 & 1 \end{vmatrix} = (-2)(1) - (-2)(2) = -2 + 4 = 2\)
3. \(\begin{vmatrix} -2 & 4 \\ 2 & 3 \end{vmatrix} = (-2)(3) - (4)(2) = -6 - 8 = -14\)

Substituting these values back into the determinant:
\[ \lambda(10) - \mu(2) + 4(-14) = 0 \]
\[ 10\lambda - 2\mu - 56 = 0 \]
\[ 5\lambda - \mu = 28 \quad \text{(Equation 1)} \]

### Step 3: Use the projection condition
The projection of \(\vec{a}\) on \(\vec{b}\) is given as \(\sqrt{54}\). The formula for the projection is:
\[ \text{Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \]

First, calculate \(|\vec{b}|\):
\[ |\vec{b}| = \sqrt{(-2)^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24} \]

Now, calculate \(\vec{a} \cdot \vec{b}\):
\[ \vec{a} \cdot \vec{b} = \lambda(-2) + \mu(4) + 4(-2) = -2\lambda + 4\mu - 8 \]

Setting up the equation using the projection:
\[ \frac{-2\lambda + 4\mu - 8}{\sqrt{24}} = \sqrt{54} \]

Cross-multiplying gives:
\[ -2\lambda + 4\mu - 8 = \sqrt{54} \cdot \sqrt{24} \]

Calculating \(\sqrt{54} \cdot \sqrt{24} = \sqrt{1296} = 36\):
\[ -2\lambda + 4\mu - 8 = 36 \]
\[ -2\lambda + 4\mu = 44 \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations
Now we have two equations:
1. \(5\lambda - \mu = 28\)
2. \(-2\lambda + 4\mu = 44\)

From Equation 1, express \(\mu\):
\[ \mu = 5\lambda - 28 \]

Substituting this into Equation 2:
\[ -2\lambda + 4(5\lambda - 28) = 44 \]
\[ -2\lambda + 20\lambda - 112 = 44 \]
\[ 18\lambda - 112 = 44 \]
\[ 18\lambda = 156 \]
\[ \lambda = 8.67 \]

Now substitute \(\lambda\) back to find \(\mu\):
\[ \mu = 5(8.67) - 28 = 43.35 - 28 = 15.35 \]

### Step 5: Find \(\lambda + \mu\)
Now, calculate \(\lambda + \mu\):
\[ \lambda + \mu = 8.67 + 15.35 = 24 \]

### Final Answer
The sum of all possible values of \(\lambda + \mu\) is:
\[ \boxed{24} \]