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]

Given vectors $$PQ = a\widehat i + b\widehat j$$ and $$PS = a\widehat i - b\widehat j$$ are adjacent sides of a parallelogram PQRS, so area of parallelogram PQRS =

|PQ $$ \times $$ PS| = 2ab = 8 (given)

$$ \therefore $$ ab = 4 ......(i)

According to the question,

|u| = |projection vector of w = $$\widehat i$$ + $$\widehat j$$ along PQ|

$$ = \left| {{{(\widehat i + \widehat j)\,.\,(a\widehat i + b\widehat j)} \over {\sqrt {{a^2} + {b^2}} }}} \right| = {{|a + b|} \over {\sqrt {{a^2} + {b^2}} }}$$

and, similarly, |v| = |projection vector of $$w = \widehat i + \widehat j$$ along PS|

$$ = \left| {{{(\widehat i + \widehat j)\,.\,(a\widehat i + b\widehat j)} \over {\sqrt {{a^2} + {b^2}} }}} \right|$$

$$ = {{|a - b|} \over {\sqrt {{a^2} + {b^2}} }} \Rightarrow |u| + |v|\, = \,|w|$$

$$ \Rightarrow {{|a + b|} \over {\sqrt {{a^2} + {b^2}} }} + {{|a - b|} \over {\sqrt {{a^2} + {b^2}} }} = \sqrt 2 $$

$$ \Rightarrow |a + b| + |a - b| = \sqrt 2 \sqrt {{a^2} + {b^2}} $$

If a $$ \ge $$ b $$ \ge $$ 0, then $$2a = \sqrt 2 \sqrt {{a^2} + {b^2}} $$

$$ \Rightarrow 4{a^2} = 2{a^2} + 2{b^2} \Rightarrow {a^2} = {b^2}$$

$$ \Rightarrow a = b$$ ....(ii)

From Eqs. (i) and (ii), we get

$$a = 2 = b \Rightarrow a + b = 4$$

and the length of diagonal PR

$$ = |a\widehat i + b\widehat j + a\widehat i - b\widehat j|$$

$$ = |2a\widehat i| = 2a = 4$$

And, the angle bisector of vector PQ and PS is along the vector

$$ \pm {\lambda \over {\sqrt {{a^2} + {b^2}} }}((a\widehat i + b\widehat j) \pm (a\widehat i - b\widehat j))$$

Since, the point A(3, 2, $$-$$1) is the mirror image of the point B(1, 0, $$-$$1) with respect to the plane $$\alpha $$x + $$\beta $$y + $$\gamma $$z = $$\delta $$, then

$${\alpha \over {3 - 1}} = {\beta \over {2 - 0}} = {\gamma \over {( - 1) - ( - 1)}}$$

$$ \Rightarrow {\alpha \over 1} = {\beta \over 1} = {\gamma \over 0} \Rightarrow \gamma = 0$$

it is given that $$\alpha + \gamma = 1 \Rightarrow \alpha = 1$$, so $$\beta = 1$$.

And, the mid-point of AB, M(2, 1, $$-$$1) lies on the given plane, so

$$2\alpha + \beta - \gamma = \delta $$

$$ \Rightarrow 2(1) + (1) - 0 = \delta \Rightarrow \delta = 3$$

$$ \therefore $$ $$\alpha + \beta = 2$$, $$\delta - \gamma = 0$$, $$\delta + \beta = 4$$.

Equation of given straight lines

$${L_1}:{{x - 1} \over 1} = {y \over { - 1}} = {{z - 1} \over 3}$$

and $${L_2}:{{x - 1} \over { - 3}} = {y \over { - 1}} = {{z - 1} \over 1}$$

having vector form respectively, are

and $$\eqalign{ & r = (\widehat i + \widehat k) + \lambda (\widehat i - \widehat j + 3\widehat k) \cr & r = (\widehat i + \widehat k) + v( - 3\widehat i - \widehat j + \widehat k) \cr} $$

$$ \because $$ $$(\widehat i - \widehat j + 3\widehat k)\,.\,( - 3\widehat i - \widehat j + \widehat k)$$

= $$-$$3 + 1 + 3 = 1 is positive,

$$ \therefore $$ Angle between supporting line vectors of lines L₁ and L₂ is acute, and point of intersection of given lines L₁ and L₂ is (1, 0, 1)

Now, vector along the acute angle bisector of vectors $$(\widehat i - \widehat j + 3\widehat k)$$ and $$( - 3\widehat i - \widehat j + \widehat k)$$ is $$(\widehat i - \widehat j + 2\widehat k)$$ or $$(\widehat i + \widehat j - 2\widehat k)$$

It is given that line

$$L:{{x - \alpha } \over l} = {{y - 1} \over m} = {{z - \gamma } \over { - 2}}$$ is the bisector of the acute angle between the lines L₁ and L₂, so

l = 1 and m = 1

and $${{I - \alpha } \over 1} = {{0 - 1} \over 1} = {{1 - \gamma } \over { - 2}}$$

$$ \Rightarrow \alpha = 2,\,\gamma = - 1$$

$$ \therefore $$ $$\alpha - \gamma = 3,\,l + m = 2$$