If $$L_1$$ is the line of intersection of the planes $$2x - 2y + 3z - 2 = 0$$, $$x - y + z + 1 = 0$$ and $$L_2$$ is the line of intersection of the planes $$x + 2y - z - 3 = 0$$, $$3x - y + 2z - 1 = 0$$, then the distance of the origin from the plane, containing the lines $$L_1$$ and $$L_2$$ is:

First, we observe that the line $$L_1$$ is obtained by the simultaneous intersection of the two planes

$$2x-2y+3z-2=0 \quad\text{and}\quad x-y+z+1=0.$$

For any line that is the intersection of two planes, its direction vector is perpendicular to the normals of both planes. The normal of the first plane is $$\mathbf n_1=(2,-2,3)$$ and the normal of the second plane is $$\mathbf n_2=(1,-1,1).$$ Hence, the direction vector $$\mathbf d_1$$ of $$L_1$$ is their cross product.

Using the determinant expansion, we have

$$\mathbf d_1=\mathbf n_1\times\mathbf n_2 =\begin{vmatrix} \mathbf i & \mathbf j & \mathbf k\\ 2 & -2 & 3\\ 1 & -1 & 1 \end{vmatrix} =\mathbf i((-2)(1)-3(-1))-\mathbf j(2\cdot1-3\cdot1)+\mathbf k(2(-1)-(-2)\cdot1) =\mathbf i( -2+3)-\mathbf j(2-3)+\mathbf k(-2+2) =(1,1,0).$$

Thus $$\mathbf d_1=(1,1,0).$$

To obtain a concrete point on $$L_1,$$ we solve the two plane equations simultaneously. Let us set $$z=t.$$ Then

$$2x-2y=2-3t,\qquad x-y=-1-t.$$

Multiplying the second relation by $$2$$ gives $$2x-2y=-2-2t.$$ Equating with the first relation,

$$-2-2t=2-3t\;\Longrightarrow\;t=4,$$

so $$z=4.$$ Substituting $$t=4$$ into $$x-y=-1-t$$ provides

$$x-y=-5.$$

Choosing $$y=0$$ yields $$x=-5.$$ Therefore a convenient point on $$L_1$$ is

$$A(-5,0,4).$$

Proceeding in exactly the same manner for $$L_2,$$ which is the intersection of

$$x+2y-z-3=0 \quad\text{and}\quad 3x-y+2z-1=0,$$

we note that the normals of these planes are $$\mathbf n_3=(1,2,-1)$$ and $$\mathbf n_4=(3,-1,2).$$ Hence the direction vector $$\mathbf d_2$$ of $$L_2$$ is

$$\mathbf d_2=\mathbf n_3\times\mathbf n_4 =\begin{vmatrix} \mathbf i & \mathbf j & \mathbf k\\ 1 & 2 & -1\\ 3 & -1 & 2 \end{vmatrix} =\mathbf i(2\cdot2-(-1)(-1))-\mathbf j(1\cdot2-(-1)\cdot3)+\mathbf k(1(-1)-2\cdot3) =(3,-5,-7).$$

Now, to find a point on $$L_2,$$ let us put $$z=s.$$ Then the two plane equations give

$$x+2y=3+s,\qquad 3x-y=1-2s.$$

From the first equation $$x=3+s-2y,$$ and substituting into the second gives

$$3(3+s-2y)-y=1-2s \;\Longrightarrow\;9+3s-6y-y=1-2s \;\Longrightarrow\;-7y=-8-5s \;\Longrightarrow\;y=\frac{8+5s}{7}.$$

Putting this back,

$$x=3+s-2\bigl(\tfrac{8+5s}{7}\bigr) =\frac{21+7s-16-10s}{7} =\frac{5-3s}{7}.$$

Choosing the simplest value $$s=0$$ gives the point

$$B\!\Bigl(\frac57,\frac87,0\Bigr).$$

The plane that contains both $$L_1$$ and $$L_2$$ must pass through the two distinct points $$A$$ and $$B$$ and must contain both direction vectors $$\mathbf d_1$$ and $$\mathbf d_2.$$ Consequently, its normal vector $$\mathbf N$$ is perpendicular to both $$\mathbf d_1$$ and $$\mathbf d_2,$$ i.e.

$$\mathbf N=\mathbf d_1\times\mathbf d_2.$$

Computing this cross product,

$$\mathbf N=\begin{vmatrix} \mathbf i & \mathbf j & \mathbf k\\ 1 & 1 & 0\\ 3 & -5 & -7 \end{vmatrix} =\mathbf i(1\cdot(-7)-0\cdot(-5)) -\mathbf j(1\cdot(-7)-0\cdot3) +\mathbf k(1\cdot(-5)-1\cdot3) =(-7,\,7,\,-8).$$

Multiplying by $$-1$$ for convenience, we take the normal as $$\mathbf N=(7,-7,8).$$

The vector equation of a plane with normal $$\mathbf N=(A,B,C)$$ passing through a point $$(x_0,y_0,z_0)$$ is

$$A(x-x_0)+B(y-y_0)+C(z-z_0)=0.$$

Substituting $$\mathbf N=(7,-7,8)$$ and point $$A(-5,0,4)$$ we get

$$7(x+5)-7(y-0)+8(z-4)=0.$$

Expanding term by term,

$$7x+35-7y+8z-32=0 \;\Longrightarrow\;7x-7y+8z+3=0.$$

This is the required plane containing both $$L_1$$ and $$L_2.$}

Finally, the distance of the origin $$O(0,0,0)$$ from the plane $$Ax+By+Cz+D=0$$ is given by the standard formula

$$\text{Distance}=\frac{|A\cdot0+B\cdot0+C\cdot0+D|}{\sqrt{A^2+B^2+C^2}}.$$

For our plane $$7x-7y+8z+3=0,$$ we have $$A=7,$$ $$B=-7,$$ $$C=8,$$ $$D=3.$$ Therefore,

$$\text{Distance}=\frac{|3|}{\sqrt{7^2+(-7)^2+8^2}} =\frac{3}{\sqrt{49+49+64}} =\frac{3}{\sqrt{162}} =\frac{3}{9\sqrt2} =\frac1{3\sqrt2}.$$

Hence, the correct answer is Option C.