Let the lines $$L_{1}:\overrightarrow{r}=\widehat{i}+2\widehat{j}+3\widehat{k}+\lambda(2\widehat{i}+3\widehat{j}+4\widehat{k}),\lambda \in \mathbb{R}$$ and $$L_{2}:\overrightarrow{r}=(4\widehat{i}+\widehat{j})+\mu (5\widehat{i}+2\widehat{j}+\widehat{k}),\mu \in \mathbb{R}$$, interest at the point R. Let P and Q be the points lying on lines $$L_{1}$$ and $$L_{2}$$ respectively, such that $$|\overrightarrow{PR}|=\sqrt{29}$$ and $$|\overrightarrow{PQ}|=\sqrt{\frac{47}{3}}$$. If the point P lies in the first octant, then $$27(QR)^{2}$$ is equal to

The lines are given as:

$$L_1: \overrightarrow{r} = \widehat{i} + 2\widehat{j} + 3\widehat{k} + \lambda (2\widehat{i} + 3\widehat{j} + 4\widehat{k})$$

$$L_2: \overrightarrow{r} = (4\widehat{i} + \widehat{j}) + \mu (5\widehat{i} + 2\widehat{j} + \widehat{k})$$

To find the point of intersection R, set the position vectors equal:

$$\widehat{i} + 2\widehat{j} + 3\widehat{k} + \lambda (2\widehat{i} + 3\widehat{j} + 4\widehat{k}) = 4\widehat{i} + \widehat{j} + \mu (5\widehat{i} + 2\widehat{j} + \widehat{k})$$

Equate coefficients:

For $$\widehat{i}$$: $$1 + 2\lambda = 4 + 5\mu \implies 2\lambda - 5\mu = 3$$ $$-(1)$$

For $$\widehat{j}$$: $$2 + 3\lambda = 1 + 2\mu \implies 3\lambda - 2\mu = -1$$ $$-(2)$$

For $$\widehat{k}$$: $$3 + 4\lambda = \mu \implies 4\lambda - \mu = -3$$ $$-(3)$$

Solve equations (2) and (3):

From (3): $$\mu = 4\lambda + 3$$ $$-(4)$$

Substitute into (2): $$3\lambda - 2(4\lambda + 3) = -1 \implies 3\lambda - 8\lambda - 6 = -1 \implies -5\lambda = 5 \implies \lambda = -1$$

From (4): $$\mu = 4(-1) + 3 = -1$$

Verify with (1): $$2(-1) - 5(-1) = -2 + 5 = 3$$, which holds.

Substitute $$\lambda = -1$$ into $$L_1$$ to find R:

$$\overrightarrow{r_R} = \widehat{i} + 2\widehat{j} + 3\widehat{k} + (-1)(2\widehat{i} + 3\widehat{j} + 4\widehat{k}) = -\widehat{i} - \widehat{j} - \widehat{k}$$

So R is $$(-1, -1, -1)$$.

Point P lies on $$L_1$$: $$\overrightarrow{r_P} = (1 + 2\lambda)\widehat{i} + (2 + 3\lambda)\widehat{j} + (3 + 4\lambda)\widehat{k}$$

Vector $$\overrightarrow{PR} = \overrightarrow{r_R} - \overrightarrow{r_P} = (-1 - (1 + 2\lambda), -1 - (2 + 3\lambda), -1 - (3 + 4\lambda)) = (-2 - 2\lambda, -3 - 3\lambda, -4 - 4\lambda)$$

Magnitude: $$|\overrightarrow{PR}| = \sqrt{(-2-2\lambda)^2 + (-3-3\lambda)^2 + (-4-4\lambda)^2} = |1 + \lambda| \sqrt{4 + 9 + 16} = |1 + \lambda| \sqrt{29}$$

Given $$|\overrightarrow{PR}| = \sqrt{29}$$, so $$|1 + \lambda| \sqrt{29} = \sqrt{29} \implies |1 + \lambda| = 1$$

Thus, $$1 + \lambda = 1$$ or $$1 + \lambda = -1 \implies \lambda = 0$$ or $$\lambda = -2$$

P is in the first octant, so coordinates positive:

If $$\lambda = 0$$: P is $$(1, 2, 3)$$

If $$\lambda = -2$$: P is $$(-3, -4, -5)$$ (not in first octant)

So P is $$(1, 2, 3)$$.

Point Q lies on $$L_2$$: $$\overrightarrow{r_Q} = (4 + 5\mu)\widehat{i} + (1 + 2\mu)\widehat{j} + \mu \widehat{k}$$

Vector $$\overrightarrow{PQ} = \overrightarrow{r_Q} - \overrightarrow{r_P} = (4 + 5\mu - 1, 1 + 2\mu - 2, \mu - 3) = (3 + 5\mu, -1 + 2\mu, \mu - 3)$$

Magnitude: $$|\overrightarrow{PQ}| = \sqrt{(3 + 5\mu)^2 + (-1 + 2\mu)^2 + (\mu - 3)^2} = \sqrt{\frac{47}{3}}$$

So: $$(3 + 5\mu)^2 + (-1 + 2\mu)^2 + (\mu - 3)^2 = \frac{47}{3}$$

Expand: $$(9 + 30\mu + 25\mu^2) + (1 - 4\mu + 4\mu^2) + (\mu^2 - 6\mu + 9) = 30\mu^2 + 20\mu + 19$$

Set equal: $$30\mu^2 + 20\mu + 19 = \frac{47}{3}$$

Multiply by 3: $$90\mu^2 + 60\mu + 57 = 47 \implies 90\mu^2 + 60\mu + 10 = 0$$

Divide by 10: $$9\mu^2 + 6\mu + 1 = 0$$

Discriminant: $$6^2 - 4 \cdot 9 \cdot 1 = 0$$, so $$\mu = -\frac{6}{18} = -\frac{1}{3}$$

Thus, Q is: $$\left(4 + 5\left(-\frac{1}{3}\right), 1 + 2\left(-\frac{1}{3}\right), -\frac{1}{3}\right) = \left(\frac{7}{3}, \frac{1}{3}, -\frac{1}{3}\right)$$

Now, QR is the distance between Q and R:

Vector $$\overrightarrow{QR} = \overrightarrow{r_R} - \overrightarrow{r_Q} = \left(-1 - \frac{7}{3}, -1 - \frac{1}{3}, -1 - \left(-\frac{1}{3}\right)\right) = \left(-\frac{10}{3}, -\frac{4}{3}, -\frac{2}{3}\right)$$

Magnitude squared: $$(QR)^2 = \left(-\frac{10}{3}\right)^2 + \left(-\frac{4}{3}\right)^2 + \left(-\frac{2}{3}\right)^2 = \frac{100}{9} + \frac{16}{9} + \frac{4}{9} = \frac{120}{9} = \frac{40}{3}$$

Now compute $$27(QR)^2 = 27 \times \frac{40}{3} = 9 \times 40 = 360$$

Therefore, the answer is 360.