Let $$\overrightarrow{OP} = \frac{\alpha - 1}{\alpha}\hat{i} + \hat{j} + \hat{k}$$, $$\overrightarrow{OQ} = \hat{i} + \frac{\beta - 1}{\beta}\hat{j} + \hat{k}$$ and $$\overrightarrow{OR} = \hat{i} + \hat{j} + \frac{1}{2}\hat{k}$$ be three vectors, where $$\alpha, \beta \in \mathbb{R} - \{0\}$$ and $$O$$ denotes the origin. If $$(\overrightarrow{OP} \times \overrightarrow{OQ}) \cdot \overrightarrow{OR} = 0$$ and the point $$(\alpha, \beta, 2)$$ lies on the plane $$3x + 3y - z + l = 0$$, then the value of $$l$$ is ________.

The components of the three position vectors are
$$\overrightarrow{OP} = \left( \frac{\alpha-1}{\alpha},\, 1,\, 1 \right),\; \overrightarrow{OQ} = \left( 1,\, \frac{\beta-1}{\beta},\, 1 \right),\; \overrightarrow{OR} = \left( 1,\, 1,\, \frac12 \right).$$

For any three vectors $$\mathbf{a},\mathbf{b},\mathbf{c}$$ originating from one point, the scalar triple product is the determinant of the matrix formed by their components:
$$(\mathbf{a}\times\mathbf{b})\cdot\mathbf{c}= \begin{vmatrix} a_x & a_y & a_z \\ b_x & b_y & b_z \\ c_x & c_y & c_z \end{vmatrix}.$$

Using $$\mathbf{a}=\overrightarrow{OP},\;\mathbf{b}=\overrightarrow{OQ},\;\mathbf{c}=\overrightarrow{OR}$$ we get

$$ \begin{vmatrix} \dfrac{\alpha-1}{\alpha} & 1 & 1 \\[4pt] 1 & \dfrac{\beta-1}{\beta} & 1 \\[4pt] 1 & 1 & \dfrac12 \end{vmatrix}=0.$$ Denote $$A = 1-\frac1{\alpha},\qquad B = 1-\frac1{\beta};$$ then the determinant becomes

$$ \begin{vmatrix} A & 1 & 1 \\ 1 & B & 1 \\ 1 & 1 & \dfrac12 \end{vmatrix}=0.$$ Expanding along the first row:

$$ A\!\left(B\cdot\frac12-1\cdot1\right) -\;1\!\left(1\cdot\frac12-1\cdot1\right) +\;1\!\left(1\cdot1-B\cdot1\right)=0. $$

Simplifying each term:
$$A\!\left(\frac{B}{2}-1\right)+\frac12+(1-B)=0,$$ so
$$A\!\left(\frac{B}{2}-1\right)+\frac32-B=0.$$

Substitute $$A=1-\dfrac1\alpha,\;B=1-\dfrac1\beta$$:

$$ \left(1-\frac1\alpha\right)\!\left(\frac12-\frac1{2\beta}-1\right)+\frac32-\left(1-\frac1\beta\right)=0. $$ First bracket: $$\frac12-\frac1{2\beta}-1=-\frac12-\frac1{2\beta}.$$

Hence
$$-\left(1-\frac1\alpha\right)\!\left(\frac12+\frac1{2\beta}\right)+\frac12+\frac1\beta=0.$$

Multiply throughout by $$2\beta$$ to clear denominators:

$$-\left(1-\frac1\alpha\right)(\beta+1)+\beta+2=0.$$

Expand:
$$-(\beta+1)+\frac{\beta+1}{\alpha}+\beta+2=0.$$

Combining like terms:
$$1+\frac{\beta+1}{\alpha}=0\;\;\Longrightarrow\;\;\frac{\beta+1}{\alpha}=-1 \;\;\Longrightarrow\;\;\beta=-\alpha-1.$$

The point $$(\alpha,\beta,2)$$ lies on the plane $$3x+3y-z+l=0$$, so

$$3\alpha+3\beta-2+l=0.$$

Using $$\beta=-\alpha-1$$:
$$3\alpha+3(-\alpha-1)-2+l=0 \;\Longrightarrow\;-5+l=0 \;\Longrightarrow\;l=5.$$

Therefore, the required value is 5.