Given $$\overrightarrow \alpha = 3\widehat i + \widehat j$$

$$\overrightarrow \beta = 2\widehat i - \widehat j + 3 \widehat k$$

$${\overrightarrow \beta _1}$$ is parallel to $$\overrightarrow \alpha $$

$$ \therefore $$ $${\overrightarrow \beta _1}$$ = $$\lambda $$ $$\overrightarrow \alpha$$

$$ \Rightarrow $$ $${\overrightarrow \beta _1}$$ = $$3\lambda \widehat i + \lambda \widehat j$$

$$\overrightarrow {{\beta _2}} $$ is perpendicular to $$\overrightarrow \alpha $$

$$\overrightarrow {{\beta _2}} $$ . $$\overrightarrow \alpha$$ = 0

Let $$\overrightarrow {{\beta _2}} $$ = $$x\widehat i + y\widehat j + z\widehat k$$

$$ \therefore $$ ($$x\widehat i + y\widehat j + z\widehat k$$).($$3\widehat i + \widehat j$$) = 0

$$ \Rightarrow $$ 3x + y = 0

$$ \Rightarrow $$ y = -3x

$$ \therefore $$ $$\overrightarrow {{\beta _2}} $$ = $$x\widehat i -3x\widehat j + z\widehat k$$

Given $$\overrightarrow \beta = {\overrightarrow \beta _1} - \overrightarrow {{\beta _2}} $$

$$ \Rightarrow $$ $$(2\widehat i - \widehat j + 3 \widehat k$$) = ($$3\lambda \widehat i + \lambda \widehat j$$) - ($$x\widehat i -3x\widehat j + z\widehat k$$)

= $$\left( {3\lambda - x} \right)\widehat i + \left( {\lambda + 3x} \right)\widehat j - z\widehat k$$

$$ \therefore $$ 3$$\lambda $$ - x = 2 ...(1)

$$\lambda $$ + 3x = -1.....(2)

z = -3

Solving (1) and (2), we get

$$\lambda $$ = $${1 \over 2}$$ and x = $$-{1 \over 2}$$

$$ \therefore $$ $${\overrightarrow \beta _1}$$ = $${3 \over 2}\widehat i + {1 \over 2}\widehat j$$

and $$\overrightarrow {{\beta _2}} $$ = $$ - {1 \over 2}\widehat i + {3 \over 2}\widehat j - 3\widehat k$$

$${\overrightarrow \beta _1} \times {\overrightarrow \beta _2} = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr {{3 \over 2}} & {{1 \over 2}} & 0 \cr { - {1 \over 2}} & {{3 \over 2}} & { - 3} \cr } } \right|$$

= $${1 \over 2}$$($$ - 3\widehat i + 9\widehat j + 5\widehat k$$)