$${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1 \Rightarrow {\cos ^2}\alpha = {1 \over 3} \Rightarrow \cos \alpha = {1 \over {\sqrt 3 }}$$

$$\overrightarrow a = {\lambda \over 3}(\widehat i + \widehat j + \widehat k),\,\lambda > 0$$

$${\lambda \over {\sqrt 3 }}{{(\widehat i + \widehat j + \widehat k)\,.\,(3\widehat i + 4\widehat j)} \over {\sqrt {{3^2} + {4^2}} }} = 7$$

$$ \Rightarrow {\lambda \over {\sqrt 3 }}(3 + 4) = 7 \times 5$$

$$\therefore$$ $$\lambda = 5\sqrt 3 $$

$$\overrightarrow a = 5(\widehat i + \widehat j + \widehat k)$$

Let $$\overrightarrow b = p\widehat i + q\widehat j + r\widehat k$$

$$\overrightarrow a \,.\,\overrightarrow b = 0$$ and $$[\overrightarrow a \,\overrightarrow b \,\widehat i] = 0$$

$$ \Rightarrow p + q + r = 0$$ ..... (i)

& $$\left| {\matrix{ p & q & r \cr 1 & 1 & 1 \cr 1 & 0 & 0 \cr } } \right| = 0 \Rightarrow \matrix{ {q = r} \cr {p = - 2r} \cr } $$

$$\overrightarrow b = - 2r\widehat i + r\widehat j + r\widehat k$$

$$\overrightarrow b = r( - 2\widehat i + \widehat j + \widehat k)$$

Now $$\left| {\overrightarrow a } \right| = \left| {\overrightarrow b } \right|$$

$$5\sqrt 3 = \left| r \right|\sqrt b \Rightarrow \left| r \right| = {5 \over {\sqrt 2 }}$$

$$\Rightarrow$$ Projection of $$\overrightarrow b $$ on $$3\widehat i + 4\widehat j = \left| {{{\overrightarrow b \,.\,\left( {3\widehat i + 4\widehat j} \right)} \over {\sqrt {{3^2} + {4^2}} }}} \right|$$

$$ = \left| r \right|{{( - 6 + 4)} \over 5} = \left| {{{ - 2r} \over 5}} \right|$$

Projection $$ = {2 \over 5} \times {5 \over {\sqrt 2 }} = \sqrt 2 $$

$$\therefore$$ B is correct.