Given two straight lines whose equations are
`(x-3)/1=(y-5)/(-2)=(z-7)/1` and `(x+1)/7=(y+1)/(-6)=(z+1)/1`
Statement 1: The line of shortest distance between the given lines is perpendicular to the plane `x+3y+5z=0`.
Statement 2 : The direction ratios of the normal to the plane `ax+by+cz+d=0` are proportonal to `a/d,b/d,c/d`.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Explanation

To solve the problem, we need to analyze the two given straight lines and the statements about them.

### Step 1: Identify the direction ratios of the two lines.

The equations of the lines are given in symmetric form:

1. For the first line:
\[ \frac{x-3}{1} = \frac{y-5}{-2} = \frac{z-7}{1} \]
The direction ratios (dr) of the first line can be taken as:
\[ b_1 = (1, -2, 1) \]

2. For the second line:
\[ \frac{x+1}{7} = \frac{y+1}{-6} = \frac{z+1}{1} \]
The direction ratios (dr) of the second line can be taken as:
\[ b_2 = (7, -6, 1) \]

### Step 2: Find the cross product of the direction ratios.

To find the line of shortest distance between the two lines, we need to compute the cross product \( b_1 \times b_2 \):

\[ b_1 \times b_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ 7 & -6 & 1 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i} \begin{vmatrix} -2 & 1 \\ -6 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 1 \\ 7 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -2 \\ 7 & -6 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \( \hat{i} \):
\[ (-2)(1) - (-6)(1) = -2 + 6 = 4 \]

2. For \( \hat{j} \):
\[ (1)(1) - (7)(1) = 1 - 7 = -6 \quad \text{(note the negative sign in front)} \]
Thus, it becomes \( +6 \).

3. For \( \hat{k} \):
\[ (1)(-6) - (7)(-2) = -6 + 14 = 8 \]

Putting it all together, we have:
\[ b_1 \times b_2 = (4, 6, 8) \]

### Step 3: Check the perpendicularity to the given plane.

The equation of the plane is:
\[ x + 3y + 5z = 0 \]
The normal vector to this plane is given by the coefficients of \( x, y, z \), which is:
\[ n = (1, 3, 5) \]

### Step 4: Check if the line of shortest distance is perpendicular to the plane.

To check if the line of shortest distance (given by the cross product \( b_1 \times b_2 \)) is perpendicular to the plane, we need to see if the direction ratios \( (4, 6, 8) \) are proportional to the normal vector \( (1, 3, 5) \).

We check the ratios:
\[ \frac{4}{1}, \frac{6}{3}, \frac{8}{5} \]
Calculating these:
- \( \frac{4}{1} = 4 \)
- \( \frac{6}{3} = 2 \)
- \( \frac{8}{5} = 1.6 \)

Since \( 4 \neq 2 \neq 1.6 \), the direction ratios are not proportional, which means the line of shortest distance is not perpendicular to the plane.

### Conclusion

- **Statement 1**: False (the line of shortest distance is not perpendicular to the plane).
- **Statement 2**: True (the direction ratios of the normal to the plane are proportional to the coefficients).