Let $$\vec{a} = \hat{i} + 5\hat{j} + \alpha\hat{k}$$, $$\vec{b} = \hat{i} + 3\hat{j} + \beta\hat{k}$$ and $$\vec{c} = -\hat{i} + 2\hat{j} - 3\hat{k}$$ be three vectors such that, $$|\vec{b} \times \vec{c}| = 5\sqrt{3}$$ and $$\vec{a}$$ is perpendicular to $$\vec{b}$$. Then the greatest amongst the values of $$|\vec{a}|^2$$ is _________.

We have the vectors $$\vec a = \hat i + 5\hat j + \alpha \hat k,\; \vec b = \hat i + 3\hat j + \beta \hat k,\; \vec c = -\hat i + 2\hat j - 3 \hat k.$$

The magnitude of the cross-product $$\vec b \times \vec c$$ is given to be $$5\sqrt3,$$ and we also know that $$\vec a$$ is perpendicular to $$\vec b.$$ We shall translate these two statements into algebraic equations step by step.

First, recall the formula for a cross-product of two vectors expressed component-wise: if $$\vec u = (u_1,u_2,u_3)$$ and $$\vec v = (v_1,v_2,v_3),$$ then $$ \vec u \times \vec v = \bigl(u_2v_3 - u_3v_2,\; u_3v_1 - u_1v_3,\; u_1v_2 - u_2v_1\bigr). $$

Applying this to $$\vec b = (1,3,\beta)$$ and $$\vec c = (-1,2,-3),$$ we obtain

$$ \vec b \times \vec c = \Bigl(3(-3) - \beta(2),\; \beta(-1) - 1(-3),\; 1\cdot 2 - 3(-1)\Bigr). $$

Simplifying each component one by one, we get

$$ \vec b \times \vec c = \bigl(-9 - 2\beta,\; -\beta + 3,\; 2 + 3\bigr) = \bigl(-9 - 2\beta,\; 3 - \beta,\; 5\bigr). $$

The squared magnitude of any vector $$\vec v = (v_1,v_2,v_3)$$ is $$|\vec v|^2 = v_1^2 + v_2^2 + v_3^2.$$ Therefore,

$$ \bigl|\vec b \times \vec c\bigr|^2 = (-9 - 2\beta)^2 + (3 - \beta)^2 + 5^2. $$

Expanding each square carefully, we write

$$ (-9 - 2\beta)^2 = (2\beta + 9)^2 = 4\beta^2 + 36\beta + 81, $$ $$ (3 - \beta)^2 = \beta^2 - 6\beta + 9, $$ $$ 5^2 = 25. $$

Adding these three expressions, we obtain

$$ |\vec b \times \vec c|^2 = 4\beta^2 + 36\beta + 81 \;+\; \beta^2 - 6\beta + 9 \;+\; 25 = 5\beta^2 + 30\beta + 115. $$

The question states that $$|\vec b \times \vec c| = 5\sqrt3,$$ so we square both sides to match the expression we have just derived:

$$ |\vec b \times \vec c|^2 = (5\sqrt3)^2 = 25\cdot3 = 75. $$

Equating the two results,

$$ 5\beta^2 + 30\beta + 115 = 75. $$

Subtracting $$75$$ from both sides, we simplify to

$$ 5\beta^2 + 30\beta + 40 = 0. $$

Dividing by $$5$$ throughout gives the quadratic

$$ \beta^2 + 6\beta + 8 = 0. $$

This factors immediately:

$$ (\beta + 2)(\beta + 4) = 0, $$ so $$ \beta = -2 \quad\text{or}\quad \beta = -4. $$

Now we employ the fact that $$\vec a$$ is perpendicular to $$\vec b.$$ The scalar (dot) product formula is

$$ \vec a \cdot \vec b = a_1b_1 + a_2b_2 + a_3b_3. $$

Writing out each component explicitly, we get

$$ \vec a \cdot \vec b = (1)(1) + (5)(3) + (\alpha)(\beta) = 1 + 15 + \alpha\beta = 16 + \alpha\beta. $$

Because the vectors are perpendicular, their dot product must vanish, hence

$$ 16 + \alpha\beta = 0 \quad\Longrightarrow\quad \alpha\beta = -16 \quad\Longrightarrow\quad \alpha = -\frac{16}{\beta}. $$

We now examine each allowed value of $$\beta$$ separately.

Case 1: $$\beta = -2$$. Then $$ \alpha = -\frac{16}{-2} = 8. $$ The squared magnitude of $$\vec a$$ is $$ |\vec a|^2 = 1^2 + 5^2 + \alpha^2 = 1 + 25 + 8^2 = 26 + 64 = 90. $$

Case 2: $$\beta = -4$$. Then $$ \alpha = -\frac{16}{-4} = 4. $$ Now $$ |\vec a|^2 = 1 + 25 + 4^2 = 26 + 16 = 42. $$

Comparing the two possibilities, the greater value of $$|\vec a|^2$$ is clearly $$90.$$

So, the answer is $$90.$$