If the shortest distance between the lines $$\vec{r} = (-\hat{i} + 3\hat{k}) + \lambda(\hat{i} - a\hat{j})$$ and $$\vec{r} = (-\hat{j} + 2\hat{k}) + \mu(\hat{i} - \hat{j} + \hat{k})$$ is $$\sqrt{\frac{2}{3}}$$, then the integral value of $$a$$ is equal to ______

We need the shortest distance between the lines:

$$L_1: \vec{r} = (-\hat{i} + 3\hat{k}) + \lambda(\hat{i} - a\hat{j})$$

$$L_2: \vec{r} = (-\hat{j} + 2\hat{k}) + \mu(\hat{i} - \hat{j} + \hat{k})$$

The direction vectors are $$\vec{d_1} = (1, -a, 0)$$ and $$\vec{d_2} = (1, -1, 1)$$. A point on $$L_1$$ is $$(-1, 0, 3)$$ and a point on $$L_2$$ is $$(0, -1, 2)$$, so

$$\vec{P_1P_2} = (0 - (-1),\, -1 - 0,\, 2 - 3) = (1, -1, -1)$$.

Next, we compute the cross product:

$$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -a & 0 \\ 1 & -1 & 1 \end{vmatrix}$$

$$= \hat{i}(-a - 0) - \hat{j}(1 - 0) + \hat{k}(-1 + a) = (-a, -1, a-1)$$.

Since the distance is given by the projection of $$\vec{P_1P_2}$$ onto the normal to both lines, we compute

$$|\vec{d_1} \times \vec{d_2}|^2 = a^2 + 1 + (a-1)^2 = 2a^2 - 2a + 2$$

and

$$\vec{P_1P_2} \cdot (\vec{d_1} \times \vec{d_2}) = (1)(-a) + (-1)(-1) + (-1)(a-1) = -a + 1 - a + 1 = 2 - 2a$$.

Thus

$$d = \frac{|2 - 2a|}{\sqrt{2a^2 - 2a + 2}} = \sqrt{\frac{2}{3}}$$.

Squaring both sides gives

$$\frac{(2 - 2a)^2}{2a^2 - 2a + 2} = \frac{2}{3}$$

which can be written as

$$\frac{4(1-a)^2}{2(a^2 - a + 1)} = \frac{2}{3}$$

and hence

$$12(1-a)^2 = 4(a^2 - a + 1)$$... expanding:

$$12(1 - 2a + a^2) = 4a^2 - 4a + 4$$

$$12 - 24a + 12a^2 = 4a^2 - 4a + 4$$

$$8a^2 - 20a + 8 = 0$$

$$2a^2 - 5a + 2 = 0$$

Factorising gives $$(2a - 1)(a - 2) = 0$$, so $$a = \frac{1}{2}$$ or $$a = 2$$. Since we require an integral value, we take $$a = 2$$.

Therefore, the answer is $$2$$.