Given,

$$\overrightarrow {OA} = 2\widehat i + 2\widehat j + 2\widehat k$$

$$\overrightarrow {OB} = \widehat i - \widehat j + 2\widehat k$$

and $$\overrightarrow {OC} = {1 \over 2}(\overrightarrow {OB} - \lambda \overrightarrow {OA} )$$

Also, $$\left| {OB \times OC} \right| = 9/2$$ ..... (i)

Now, $$\overrightarrow {OB} \times \overrightarrow {OC} = \overrightarrow {OB} \times {1 \over 2}(\overrightarrow {OB} - \lambda \overrightarrow {OA} )$$

$$ = {{ - \lambda } \over 2}\overrightarrow {OB} \times \overrightarrow {OA} = {\lambda \over 2}(\overrightarrow {OA} \times \overrightarrow {OB} )$$

We need to find $$\overrightarrow {OA} \times \overrightarrow {OB} $$ for this,

($$\because$$ $$\overrightarrow a \times \overrightarrow b = - \overrightarrow b \times \overrightarrow a $$ and $$\overrightarrow a \times \overrightarrow a $$ = 0)

$$\overrightarrow {OA} \times \overrightarrow {OB} = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 2 & 1 \cr 1 & { - 2} & 2 \cr } } \right|$$

$$ = 6\widehat i - 3\widehat j - 6\widehat k = 3(2\widehat i - \widehat j - 2\widehat k)$$

So, $$\overrightarrow {OB} \times \overrightarrow {OC} = {{3\lambda } \over 2}(2\widehat i - \widehat j - 2\widehat k)$$

From Eq. (i), $$\left| {{{3\lambda } \over 2}\sqrt {4 + 1 + 4} } \right| = 9/2$$

$$ \Rightarrow \left| {{{9\lambda } \over 2}} \right| = 9/2 \Rightarrow {9 \over 2}\left| \lambda \right| = {9 \over 2}$$

$$ \Rightarrow \left| \lambda \right| = 1 \Rightarrow \lambda = \pm 1$$

But $$\lambda$$ > 0

$$\therefore$$ $$\lambda$$ = 1

So, $$\overrightarrow {OC} = {1 \over 2}(\overrightarrow {OB} - \overrightarrow {OA} )$$ (By putting $$\lambda$$ = 1)

$$ \Rightarrow \overrightarrow {OC} = {1 \over 2}( - \widehat i - 4\widehat j + \widehat k) = - {1 \over 2}\widehat i - 2\widehat j + {1 \over 2}\widehat k$$

Option (a)

Projection of $${\overrightarrow {OC} }$$ and $${\overrightarrow {OA} }$$ $$ = {{\overrightarrow {OC} \,.\,\overrightarrow {OA} } \over {\left| {\overrightarrow {OA} } \right|}}$$

$$ = {{{1 \over 2}( - 2 - 8 + 1)} \over 3} = {{ - 3} \over 2}$$

Option (b)

Area of $$\Delta OAB = {1 \over 2}\left| {\overrightarrow {OA} \times \overrightarrow {OB} } \right| = {1 \over 2}\left| {\overrightarrow {OA} } \right|\left. {\overrightarrow {OB} } \right|\sin 90^\circ $$

$$ = {1 \over 2}\left| {3 \times 3} \right| = {9 \over 2}$$

Option (c)

Area of the $$\Delta ABC = {1 \over 2}\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right|$$

$$ = {1 \over 2}\left\| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr { - 1} & { - 4} & 1 \cr { - 5/2} & { - 4} & {{{ - 1} \over 2}} \cr } } \right\|$$

$$ = {1 \over 2}\left| {6\widehat i - 3\widehat j - 6\widehat k} \right| = {1 \over 2} \times 3 \times 3 = {9 \over 2}$$

Option (d)

The acute angle between the diagonals of the parallelogram with adjacent sides $${\overrightarrow {OA} }$$ and $${\overrightarrow {OC} }$$ = $$\theta$$

$$ \Rightarrow {{(\overrightarrow {OA} + \overrightarrow {OC} )\,.\,(\overrightarrow {OA} - \overrightarrow {OC} )} \over {\left| {\overrightarrow {OA} + \overrightarrow {OC} } \right|\left| {\overrightarrow {OA} - \overrightarrow {OC} } \right|}} = \cos \theta $$

$$ \Rightarrow \cos \theta = {{\left( {{3 \over 2}\widehat i + {3 \over 2}\widehat k} \right)\,.\,\left( {{5 \over 2}\widehat i + 4\widehat j + {1 \over 2}\widehat k} \right)} \over {{3 \over 2}\sqrt 2 \times \sqrt {{{90} \over 4}} }}$$

$$ = {{18} \over {3\sqrt 2 \times \sqrt {90} }} = {1 \over {\sqrt 5 }} \ne {1 \over 2}$$ [$$\therefore$$ $$\theta$$ $$\ne$$ $$\pi$$/3]