Let $$\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}, a_i > 0, i = 1, 2, 3$$ be a vector which makes equal angles with the coordinate axes $$OX, OY$$ and $$OZ$$. Also, let the projection of $$\vec{a}$$ on the vector $$3\hat{i} + 4\hat{j}$$ be $$7$$. Let $$\vec{b}$$ be a vector obtained by rotating $$\vec{a}$$ with $$90°$$. If $$\vec{a}, \vec{b}$$ and x-axis are coplanar, then projection of a vector $$\vec{b}$$ on $$3\hat{i} + 4\hat{j}$$ is equal to

We begin by noting that $$\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$$ with each $$a_i > 0$$ and that it makes equal angles with the coordinate axes. Consequently, $$a_1 = a_2 = a_3 = k$$ for some $$k > 0$$. The projection of $$\vec{a}$$ on $$3\hat{i} + 4\hat{j}$$ is given by
$$ \frac{\vec{a} \cdot (3\hat{i} + 4\hat{j})}{|3\hat{i} + 4\hat{j}|} = \frac{3k + 4k}{5} = \frac{7k}{5} = 7 $$
which implies $$k = 5$$ and hence $$\vec{a} = 5\hat{i} + 5\hat{j} + 5\hat{k}$$.

Next, to determine $$\vec{b}$$ we observe that rotating $$\vec{a}$$ by $$90°$$ preserves its magnitude, so $$|\vec{b}| = |\vec{a}| = 5\sqrt{3}$$ while also ensuring $$\vec{a} \cdot \vec{b} = 0$$. Because $$\vec{a}$$, $$\vec{b}$$, and the x-axis are coplanar, $$\vec{b}$$ lies in the plane spanned by $$\vec{a}$$ and $$\hat{i}$$. Thus we may write
$$ \vec{b} = \alpha\vec{a} + \beta\hat{i} $$

Then, imposing the orthogonality condition $$\vec{a} \cdot \vec{b} = 0$$ yields
$$ \alpha|\vec{a}|^2 + \beta(\vec{a} \cdot \hat{i}) = 0 \implies 75\alpha + 5\beta = 0 \implies \beta = -15\alpha $$
Hence $$\vec{b} = \alpha(5,5,5) + (-15\alpha)(1,0,0) = \alpha(-10,5,5)$$. Next, enforcing the magnitude condition $$|\vec{b}| = 5\sqrt{3}$$ gives
$$ \alpha^2(100+25+25) = 75 \implies 150\alpha^2 = 75 \implies \alpha = \pm\frac{1}{\sqrt{2}} $$

Therefore
$$ \vec{b} = \pm\frac{1}{\sqrt{2}}(-10,5,5) $$
and its projection on $$3\hat{i} + 4\hat{j}$$ is
$$ \text{Projection} = \frac{\vec{b} \cdot (3\hat{i} + 4\hat{j})}{5} = \frac{\pm\frac{1}{\sqrt{2}}(-30+20)}{5} = \frac{\pm\frac{1}{\sqrt{2}}\cdot(-10)}{5} = \frac{\mp10}{5\sqrt{2}} = \mp\sqrt{2} $$
Taking the positive value yields $$\sqrt{2}$$.

The answer is Option B: $$\sqrt{2}$$.