The square of the distance of the point $$(\frac{15}{7},\frac{32}{7},7)$$ from the line $$\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$$ in the direction of the vector $$\hat{i}+4\hat{j}+7\hat{k}$$ is :

We need to find the square of the distance from the point $$P = \left(\frac{15}{7}, \frac{32}{7}, 7\right)$$ to the line $$\frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}$$, measured in the direction of $$\vec{d} = \hat{i} + 4\hat{j} + 7\hat{k}$$.

We start by parametrizing the ray from $$P$$ in the given direction. A point on the ray is: $$ Q = P + t\vec{d} = \left(\frac{15}{7}+t,\; \frac{32}{7}+4t,\; 7+7t\right) $$.

Next, the line can be parametrized as $$(x,y,z) = (-1+3s,\; -3+5s,\; -5+7s)$$. Setting $$Q$$ equal to a point on this line gives

$$ \frac{15}{7} + t = -1 + 3s \quad \cdots(1) $$

$$ \frac{32}{7} + 4t = -3 + 5s \quad \cdots(2) $$

$$ 7 + 7t = -5 + 7s \quad \cdots(3) $$

From (3) we obtain $$7s - 7t = 12 \implies s - t = \frac{12}{7} \implies s = t + \frac{12}{7} \quad \cdots(4)$$. Substituting this into (1) yields

$$\frac{15}{7} + t = -1 + 3\left(t + \frac{12}{7}\right) = -1 + 3t + \frac{36}{7}$$

$$\frac{15}{7} + t = \frac{-7+36}{7} + 3t = \frac{29}{7} + 3t$$

$$\frac{15}{7} - \frac{29}{7} = 2t$$

$$-\frac{14}{7} = 2t \implies t = -1$$

It follows from (4) that $$s = -1 + \frac{12}{7} = \frac{5}{7}$$.

As a check, substituting into (2) gives on the left $$\frac{32}{7} + 4(-1) = \frac{32}{7} - 4 = \frac{4}{7}$$ and on the right $$-3 + 5 \cdot \frac{5}{7} = -3 + \frac{25}{7} = \frac{4}{7}$$, confirming consistency.

The distance from $$P$$ to the intersection point $$Q$$ equals $$|t| \cdot |\vec{d}|$$, where $$|t| = 1$$ and $$|\vec{d}| = \sqrt{1+16+49} = \sqrt{66}$$. Therefore, the square of the distance is

$$ |t|^2 \cdot |\vec{d}|^2 = 1 \times 66 = 66 $$.

The correct answer is Option 4: 66.