$$7x + 8y + 9z - (\gamma - 1) = A(4x + 5y + 6z - \beta ) + B(x + 2y + 3z - \alpha )$$

On equating the coefficients,

4A + B = 7 .... (i)

5A + 2B = 8 .... (ii)

and $$-$$ ($$\gamma$$ $$-$$ 1) = $$-$$ A$$\beta$$ $$-$$ $$\alpha$$B ..... (iii)

On solving Eqs. (i) and (ii), we get A = 2 and B = $$-$$1

From Eq. (iii), we get

$$-$$ $$\gamma$$ + 1 = $$-$$ 2$$\beta$$ $$-$$ $$\alpha$$($$-$$1)

$$\Rightarrow$$ $$\alpha$$ $$-$$ 2$$\beta$$ + $$\gamma$$ = 1 ..... (iv)

Now, determinant of

$$M = \left| M \right| = \left| {\matrix{ \alpha & 2 & \gamma \cr \beta & 1 & 0 \cr { - 1} & 0 & 1 \cr } } \right| = \alpha - 2\beta + \gamma = 1$$ [from Eq. (iv)]

Equation of plane P is given by $$x - 2y + z = 1$$

Hence, perpendicular distance of the point (0, 1, 0) from the plane

$$P = {{\left| {0 - 2 \times 1 + 0 - 1} \right|} \over {\sqrt {{1^2} + {{( - 2)}^2} + {1^2}} }} = {{\left| 3 \right|} \over {\sqrt 6 }}$$

$$ \Rightarrow D = {\left( {{{\left| 3 \right|} \over {\sqrt 6 }}} \right)^2} = {9 \over 6} = 1.5$$

$$7x + 8y + 9z - (\gamma - 1) = A(4x + 5y + 6z - \beta ) + B(x + 2y + 3z - \alpha )$$

On equating the coefficients,

4A + B = 7 .... (i)

5A + 2B = 8 .... (ii)

and $$-$$ ($$\gamma$$ $$-$$ 1) = $$-$$ A$$\beta$$ $$-$$ $$\alpha$$B ..... (iii)

On solving Eqs. (i) and (ii), we get A = 2 and B = $$-$$1

From Eq. (iii), we get

$$-$$ $$\gamma$$ + 1 = $$-$$ 2$$\beta$$ $$-$$ $$\alpha$$($$-$$1)

$$\Rightarrow$$ $$\alpha$$ $$-$$ 2$$\beta$$ + $$\gamma$$ = 1 ..... (iv)

Now, determinant of

$$M = \left| M \right| = \left| {\matrix{ \alpha & 2 & \gamma \cr \beta & 1 & 0 \cr { - 1} & 0 & 1 \cr } } \right| = \alpha - 2\beta + \gamma = 1$$ [from Eq. (iv)]

Given vectors $$\overrightarrow a $$$$ = 2\widehat i + \widehat j - \widehat k$$

and $$\overrightarrow b = \widehat i + 2\widehat j + \widehat k$$

So, $$\overrightarrow a + \overrightarrow b = 3\widehat i + 3\widehat j \Rightarrow |\overrightarrow a + \overrightarrow b| = 3\sqrt 2 $$

Since, it is given that projection of $$\overrightarrow c $$ = $$\alpha $$a + $$\beta $$b on the vector ($$\overrightarrow a $$ + $$\overrightarrow b $$) is $$3\sqrt 2 $$, then

$${{(\overrightarrow a + \overrightarrow b ).\overrightarrow c } \over {|\overrightarrow a + \overrightarrow b|}} = 3\sqrt 2 $$

$$ \Rightarrow (\overrightarrow a + \overrightarrow b).(\alpha \overrightarrow a + \beta \overrightarrow b) = 18$$

$$ \Rightarrow \alpha (\overrightarrow a.\overrightarrow a) + \beta (\overrightarrow a.\overrightarrow b) + \alpha (\overrightarrow b.\overrightarrow a) + \beta (\overrightarrow a.\overrightarrow b) = 18$$

$$ \Rightarrow 6\alpha + 3\beta + 3\alpha + 6\beta = 18$$

$$ \Rightarrow 9\alpha + 9\beta = 18 \Rightarrow (\alpha + \beta ) = 2$$ .....(i)

Now, for minimum value of

($$\overrightarrow c$$ $$-$$($$\overrightarrow a$$ $$ \times $$ $$\overrightarrow b$$)).$$\overrightarrow c$$

$$ = (\alpha \overrightarrow a + \beta \overrightarrow b - (\overrightarrow a \times \overrightarrow b)).(\alpha \overrightarrow a + \beta \overrightarrow b)$$

$$ = {\alpha ^2}(\overrightarrow a.\overrightarrow a) + \alpha \beta (\overrightarrow a.\overrightarrow b) + \alpha \beta (\overrightarrow a.\overrightarrow b) + {\beta ^2}(\overrightarrow b.\overrightarrow b)$$

[$$ \because $$ ($$\overrightarrow a$$ $$ \times $$ $$\overrightarrow b$$) . $$\overrightarrow a$$ = 0 = ($$\overrightarrow a $$$$ \times $$ $$\overrightarrow b$$) . $$\overrightarrow b$$]

$$6{\alpha ^2} + 6\alpha \beta + 6{\beta ^2} = 6({\alpha ^2} + {\beta ^2} + \alpha \beta )$$

$$ = 6\,[{(\alpha + \beta )^2} - \alpha \beta ] = 6\,[4 - \alpha \beta ]$$

$$ = 6\,[4 - \alpha (2 - \alpha )]$$

$$ = 6\,[4 - 2\alpha + {\alpha ^2}]$$

Let f ( $$\alpha $$) = $$4 - 2\alpha + {\alpha ^2}$$

f′( $$\alpha $$) = –2 + 2$$\alpha $$

At maximum and minimum f′( $$\alpha $$) = 0

$$ \Rightarrow $$ –2 + 2$$\alpha $$ $$ \Rightarrow $$ $$\alpha $$ = 1

f′'( $$\alpha $$) = 2 (+ve)

Therefore, minimum value of $$4 - 2\alpha + {\alpha ^2}$$ is (4 – 2 + 1) = 3.

$$ \therefore $$ The minimum value of

$$6(4 - 2\alpha + {\alpha^2}) = 6(3) = 18$$