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

To find the value of $l$, we start by analyzing the given vectors:
$\overrightarrow{OP}=\frac{\alpha -1}{\alpha }\hat{i}+\hat{j}+\hat{k}$
$\overrightarrow{OQ}=\hat{i}+\frac{\beta -1}{\beta }$
$\overrightarrow{OR}=\hat{i}+\hat{j}+\frac{1}{2}$

We are given that the scalar triple product is zero:
$(\overrightarrow{OP}\times \stackrel{���}{OQ})\cdot \overrightarrow{OR}=0$
This is equivalent to stating that the determinant formed by the components of these three vectors is equal to zero:
$\left|\begin{array}{ccc}1-\frac{1}{\alpha }& 1& 1\\ 1& 1-\frac{1}{\beta }& 1\\ 1& 1& \frac{1}{2}\end{array}\right|=0$

Let us perform row operations to simplify the determinant. Apply $R_1\to R_1-R_3$ and $R_2\to R_2-R_3$:
$\left|\begin{array}{ccc}-\frac{1}{\alpha }& 0& \frac{1}{2}\\ 0& -\frac{1}{\beta }& \frac{1}{2}\\ 1& 1& \frac{1}{2}\end{array}\right|=0$

Now, expanding the determinant along the first row:
$-\frac{1}{\alpha }[-\frac{1}{2\beta }-\frac{1}{2}]+\frac{1}{2}[0-(-\frac{1}{\beta }\left)\right]=0$
Simplifying the terms:
$\frac{1}{2\alpha \beta }+\frac{1}{2\alpha }+\frac{1}{2\beta }=0$
Multiplying the entire equation by $2\alpha \beta$ (since $\alpha ,\beta \ne 0$):
$1+\beta +\alpha =0$
Thus, we obtain the relation:
$\alpha +\beta =-1$

We are given that the point $(\alpha ,\beta ,2)$ lies on the plane:
$3x+3y-z+l=0$
Substituting $x=\alpha$, $y=\beta$, and $z=2$ into the plane equation:
$3\alpha +3\beta -2+l=0$
Factor out $3$ from the first two terms:
$3(\alpha +\beta )-2+l=0$
Substitute the relation $\alpha +\beta =-1$:
$3(-1)-2+l=0$
$-3-2+l=0$
$-5+l=0$
$l=5$

Therefore, the value of $l$ is 5.