The largest value of $$𝑎$$, for which the perpendicular distance of the plane containing the lines $$\vec{r} = \hat{i} + \hat{j} + \lambda( \hat{i} + a \hat{j} - \hat{k} )\quad \text{and} \quad \vec{r} = \hat{i} + \hat{j} + \mu (\hat{i} + \hat{j} - a \hat{k})$$ from the point 2, 1, 4 is $$\sqrt{3}$$, is 

The given lines are

$$\vec r=(\hat i+\hat j)+\lambda(\hat i+a\hat j-\hat k)$$

and

$$\vec r=(\hat i+\hat j)+\mu(-\hat i+\hat j-a\hat k)$$

Both pass through the point

$$(1,1,0)$$

Their direction vectors are

$$\vec d_1=(1,a,-1)$$

and

$$\vec d_2=(-1,1,-a)$$

Hence, the normal vector to the plane containing these lines is

$$\vec n=\vec d_1\times\vec d_2$$

$$=\begin{vmatrix} \hat i&\hat j&\hat k\\ 1&a&-1\\ -1&1&-a \end{vmatrix}$$

$$=\hat i(1-a^2)-\hat j(-a+1)+\hat k(1+a)$$

$$=(1-a)\hat i+(a-1)\hat j+(1+a)\hat k$$

Taking normal vector as

$$\vec n=(1-a,1,1+a),$$

equation of plane through $$(1,1,0)$$ is

$$(1-a)(x-1)+(y-1)+(1+a)z=0$$

$$=(1-a)x+y+(1+a)z+a-2=0$$

Now distance of point $$(2,1,4)$$ from the plane is

$$\frac{|(1-a)2+1+4(1+a)+a-2|}{\sqrt{(1-a)^2+1+(1+a)^2}}=\sqrt3$$

Simplifying,

$$\frac{|5-a|}{\sqrt{a^2-2a+3}}=\sqrt3$$

Squaring both sides,

$$\frac{(5-a)^2}{a^2-2a+3}=3$$

$$a^2-10a+25=3a^2-6a+9$$

$$2a^2+4a-16=0$$

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

$$(a-2)(a+4)=0$$

Hence,

$$a=2\quad \text{or}\quad a=-4$$

Therefore, the largest value of $$a$$ is

$$\boxed{2}$$.