Let the plane $$P : \vec{r} \cdot \vec{a} = d$$ contain the line of intersection of two planes $$\vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 6$$ and $$\vec{r} \cdot (-6\hat{i} + 5\hat{j} - \hat{k}) = 7$$. If the plane $$P$$ passes through the point $$(2, 3, \frac{1}{2})$$, then the value of $$\frac{|13\vec{a}|^2}{d^2}$$ is equal to

The two given planes are:

$$P_1: \vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 6 \implies x + 3y - z = 6$$

$$P_2: \vec{r} \cdot (-6\hat{i} + 5\hat{j} - \hat{k}) = 7 \implies -6x + 5y - z = 7$$

Any plane through the line of intersection is:

$$(x + 3y - z - 6) + \lambda(-6x + 5y - z - 7) = 0$$

$$(1 - 6\lambda)x + (3 + 5\lambda)y + (-1 - \lambda)z = 6 + 7\lambda$$

This plane passes through $$(2, 3, \frac{1}{2})$$:

$$(1-6\lambda)(2) + (3+5\lambda)(3) + (-1-\lambda)\left(\frac{1}{2}\right) = 6 + 7\lambda$$

$$2 - 12\lambda + 9 + 15\lambda - \frac{1}{2} - \frac{\lambda}{2} = 6 + 7\lambda$$

$$\frac{21}{2} + \frac{5\lambda}{2} = 6 + 7\lambda$$

Multiplying by 2: $$21 + 5\lambda = 12 + 14\lambda$$

$$9 = 9\lambda \implies \lambda = 1$$

The plane $$P$$ is:

$$(1-6)x + (3+5)y + (-1-1)z = 6 + 7$$

$$-5x + 8y - 2z = 13$$

So $$\vec{a} = -5\hat{i} + 8\hat{j} - 2\hat{k}$$ and $$d = 13$$.

Now computing $$\frac{|13\vec{a}|^2}{d^2}$$:

$$|13\vec{a}|^2 = 169(25 + 64 + 4) = 169 \times 93$$

$$d^2 = 169$$

$$\frac{|13\vec{a}|^2}{d^2} = \frac{169 \times 93}{169} = 93$$

Hence the correct answer is Option B: $$93$$.