Here, P(1, 0, 0), Q(0, 1, 0), R(0, 0, 1), T = (1, 1, 1) and $$S\left( {{1 \over 2},{1 \over 2},{1 \over 2}} \right)$$.

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Now, $$\overrightarrow p = \overrightarrow {SP} = \overrightarrow {OP} - \overrightarrow {OS} $$

$$ = \left( {{1 \over 2}\widehat i - {1 \over 2}\widehat j - {1 \over 2}\widehat k} \right) = {1 \over 2}(\widehat i - \widehat j - \widehat k)$$

$$\overrightarrow q = \overrightarrow {SQ} = {1 \over 2}( - \widehat i + \widehat j - \widehat k)$$

$$\overrightarrow r = \overrightarrow {SR} = {1 \over 2}( - \widehat i - \widehat j + \widehat k)$$

and $$\overrightarrow t = \overrightarrow {ST} = {1 \over 2}(\widehat i + \widehat j + \widehat k)$$

$$\overrightarrow p \times \overrightarrow q = {1 \over 4}\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & { - 1} & { - 1} \cr { - 1} & 1 & { - 1} \cr } } \right| = {1 \over 4}(2\widehat i + 2\widehat j)$$

and $$\overrightarrow r \, \times \,\overrightarrow t = {1 \over 4}\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr { - 1} & { - 1} & 1 \cr 1 & 1 & 1 \cr } } \right| = {1 \over 4}( - 2\widehat i + 2\widehat j)$$

Now, $$(\overrightarrow p \, \times \,\overrightarrow q )\, \times \,(\overrightarrow r \, \times \,\overrightarrow t ) = {1 \over {16}}\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 2 & 0 \cr { - 2} & 2 & 0 \cr } } \right| = {1 \over {16}}(8\widehat k) = {1 \over 2}\widehat k$$

$$ \therefore $$ $$|(p \times q) \times (\overrightarrow r \, \times \,\overrightarrow t )|\, = \,\left| {{1 \over 2}\widehat k} \right| = {1 \over 2} = 0.5$$

Given,

$$f(x + y) = f(x)f'(y) + f'(x)f(y),\,\forall x,y \in R$$ and f(0) = 1

Put x = y = 0, we get

f(0) = f(0) f'(0) + f'(0) f(0)

$$ \Rightarrow 1 = 2f'(0) \Rightarrow f'(0) = {1 \over 2}$$

Put x = x and y = 0, we get

f(x) = f(x) f'(0) + f'(x) f(0)

$$ \Rightarrow f(x) = {1 \over 2}f(x) + f'(x)$$

$$ \Rightarrow f'(x) = {1 \over 2}f(x) \Rightarrow {{f'(x)} \over {f(x)}} = {1 \over 2}$$

On integrating, we get

$$\log f(x) = {1 \over 2}x + C$$

$$ \Rightarrow f(x) = A{e^{{1 \over 2}x}}$$, where e^C = A

If f(0) = 1, then A = 1

Hence, $$f(x) = {e^{{1 \over 2}x}}$$

$$ \Rightarrow {\log _e}f(x) = {1 \over 2}x$$

$$ \Rightarrow {\log _e}f(4) = {1 \over 2} \times 4 = 2$$