If the shortest distance between the lines $$\frac{x - \lambda}{2} = \frac{y - 4}{3} = \frac{z - 3}{4}$$ and $$\frac{x - 2}{4} = \frac{y - 4}{6} = \frac{z - 7}{8}$$ is $$\frac{13}{\sqrt{29}}$$, then a value of $$\lambda$$ is :

Consider the two lines given by

$$L_1: \frac{x-\lambda}{2} = \frac{y-4}{3} = \frac{z-3}{4}$$ (direction $$\vec{d_1} = (2,3,4)$$, point $$A = (\lambda,4,3)$$)

$$L_2: \frac{x-2}{4} = \frac{y-4}{6} = \frac{z-7}{8}$$ (direction $$\vec{d_2} = (4,6,8)$$, point $$B = (2,4,7)$$)

Since $$\vec{d_2} = 2\vec{d_1}$$, the two lines are parallel. For parallel lines, the shortest distance is given by

$$d = \frac{|\vec{AB} \times \vec{d_1}|}{|\vec{d_1}|}$$

Here, $$\vec{AB} = B - A = (2-\lambda, 0, 4)$$ and

$$\vec{AB} \times \vec{d_1} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2-\lambda & 0 & 4 \\ 2 & 3 & 4 \end{vmatrix}$$

$$= \hat{i}(0 \cdot 4 - 4 \cdot 3) - \hat{j}((2-\lambda) \cdot 4 - 4 \cdot 2) + \hat{k}((2-\lambda) \cdot 3 - 0 \cdot 2)$$

$$= \hat{i}(-12) - \hat{j}(8-4\lambda-8) + \hat{k}(6-3\lambda)$$

$$= -12\hat{i} + 4\lambda\hat{j} + (6-3\lambda)\hat{k}$$

It follows that

$$|\vec{AB} \times \vec{d_1}| = \sqrt{144 + 16\lambda^2 + (6-3\lambda)^2} = \sqrt{144 + 16\lambda^2 + 36 - 36\lambda + 9\lambda^2}$$

$$= \sqrt{25\lambda^2 - 36\lambda + 180}$$

and

$$|\vec{d_1}| = \sqrt{4+9+16} = \sqrt{29} \,.$$

Thus the distance is

$$d = \frac{\sqrt{25\lambda^2 - 36\lambda + 180}}{\sqrt{29}} = \frac{13}{\sqrt{29}}$$

Squaring both sides gives

$$25\lambda^2 - 36\lambda + 180 = 169$$

$$25\lambda^2 - 36\lambda + 11 = 0$$

Using the quadratic formula,

$$\lambda = \frac{36 \pm \sqrt{1296 - 1100}}{50} = \frac{36 \pm \sqrt{196}}{50} = \frac{36 \pm 14}{50}$$

$$\lambda = \frac{50}{50} = 1 \quad \text{or} \quad \lambda = \frac{22}{50} = \frac{11}{25}$$

From the options, $$\lambda = 1$$ is available.

The correct answer is Option 4: 1.