If the vectors $$\vec{a} = \lambda\hat{i} + \mu\hat{j} + 4\hat{k}$$, $$\vec{b} = -2\hat{i} + 4\hat{j} - 2\hat{k}$$ and $$\vec{c} = 2\hat{i} + 3\hat{j} + \hat{k}$$ 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

The given vectors are:

$$\vec{a} = \lambda\hat{i} + \mu\hat{j} + 4\hat{k}$$

$$\vec{b} = -2\hat{i} + 4\hat{j} - 2\hat{k}$$

$$\vec{c} = 2\hat{i} + 3\hat{j} + \hat{k}$$

Since the vectors $$\vec{a}$$, $$\vec{b}$$, and $$\vec{c}$$ are coplanar, their scalar triple product must be zero:

$$\begin{vmatrix} \lambda & \mu & 4 \\ -2 & 4 & -2 \\ 2 & 3 & 1 \end{vmatrix} = 0$$

Expanding the determinant along the first row:

$$\lambda(4(1) - (-2)(3)) - \mu((-2)(1) - (-2)(2)) + 4((-2)(3) - 4(2)) = 0$$

$$\lambda(4 + 6) - \mu(-2 + 4) + 4(-6 - 8) = 0$$

$$10\lambda - 2\mu - 56 = 0 \implies 5\lambda - \mu = 28$$

Rearranging for $$\mu$$ gives:

$$\mu = 5\lambda - 28$$

Next, the magnitude of the projection of $$\vec{a}$$ on $$\vec{b}$$ is given as $$\sqrt{54}$$ units:

$$\frac{|\vec{a} \cdot \vec{b}|}{|\vec{b}|} = \sqrt{54}$$

Computing the dot product $$\vec{a} \cdot \vec{b}$$ and the magnitude $$|\vec{b}|$$:

$$\vec{a} \cdot \vec{b} = (\lambda)(-2) + (\mu)(4) + (4)(-2) = -2\lambda + 4\mu - 8$$

$$|\vec{b}| = \sqrt{(-2)^2 + 4^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24}$$

Substitute these values into the projection equation:

$$\frac{|-2\lambda + 4\mu - 8|}{\sqrt{24}} = \sqrt{54}$$

$$|-2\lambda + 4\mu - 8| = \sqrt{54 \times 24} = \sqrt{1296} = 36$$

Dividing by 2 inside the absolute value yields:

$$|-\lambda + 2\mu - 4| = 18$$

This gives two possible cases to explore:

Case 1: $$-\lambda + 2\mu - 4 = 18 \implies -\lambda + 2\mu = 22$$

Substituting $$\mu = 5\lambda - 28$$ into this linear relation:

$$-\lambda + 2(5\lambda - 28) = 22 \implies 9\lambda = 78 \implies \lambda = \frac{26}{3}$$

$$\mu = 5\left(\frac{26}{3}\right) - 28 = \frac{46}{3}$$

Thus, the first possible value for the sum is:

$$(\lambda + \mu)_1 = \frac{26}{3} + \frac{46}{3} = \frac{72}{3} = 24$$

Case 2: $$-\lambda + 2\mu - 4 = -18 \implies -\lambda + 2\mu = -14$$

Substituting $$\mu = 5\lambda - 28$$ into this linear relation:

$$-\lambda + 2(5\lambda - 28) = -14 \implies 9\lambda = 42 \implies \lambda = \frac{14}{3}$$

$$\mu = 5\left(\frac{14}{3}\right) - 28 = -\frac{14}{3}$$

Thus, the second possible value for the sum is:

$$(\lambda + \mu)_2 = \frac{14}{3} + \left(-\frac{14}{3}\right) = 0$$

Summing all possible values of $$\lambda + \mu$$:

$$\text{Sum} = 24 + 0 = 24$$

Conclusion:

The sum of all possible values of $$\lambda + \mu$$ is equal to 24.