Statement 1: A point on the line `(x+2)/3=(y+1)/2=(z-3)/2` at a distance `3sqrt(2)` from the point `(1,2,3)` lies on the lne `(x+7)/5=(y+5)/4=(z-2)/1`
Statement 2: If `d` is the distance between the point `(-1,-5,-10)` and the point of intersectionof the line `(x-2)/3=(y+1)/4=(z-2)/12` with the plane `x-y+z=5` then `d=13`

Options

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

Explanation

To solve the problem, we will analyze both statements step by step.

### Statement 1:
We need to find a point on the line given by the equation:
\[ \frac{x + 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{2} \]
Let this common ratio be \( r \). Then we can express \( x, y, z \) in terms of \( r \):
\[ x = 3r - 2, \quad y = 2r - 1, \quad z = 2r + 3 \]

Next, we need to find the distance from the point \( (1, 2, 3) \) to the point \( (3r - 2, 2r - 1, 2r + 3) \). The distance formula in 3D space is given by:
\[ d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2} \]
Substituting the coordinates:
\[ d = \sqrt{(1 - (3r - 2))^2 + (2 - (2r - 1))^2 + (3 - (2r + 3))^2} \]
This simplifies to:
\[ d = \sqrt{(3 - 3r)^2 + (3 - 2r)^2 + (-2r)^2} \]
Calculating each term:
1. \( (3 - 3r)^2 = 9(1 - r)^2 = 9(1 - 2r + r^2) = 9 - 18r + 9r^2 \)
2. \( (3 - 2r)^2 = 9 - 12r + 4r^2 \)
3. \( (-2r)^2 = 4r^2 \)

Combining these:
\[ d^2 = (9 - 18r + 9r^2) + (9 - 12r + 4r^2) + 4r^2 \]
\[ = 18 - 30r + 22r^2 \]

We know that the distance is given as \( 3\sqrt{2} \), so:
\[ d^2 = (3\sqrt{2})^2 = 18 \]
Setting the two expressions for \( d^2 \) equal:
\[ 18 - 30r + 22r^2 = 18 \]
This simplifies to:
\[ -30r + 22r^2 = 0 \]
Factoring out \( r \):
\[ r(22r - 30) = 0 \]
Thus, \( r = 0 \) or \( r = \frac{30}{22} = \frac{15}{11} \).

Now, substituting \( r = 0 \):
\[ x = -2, \quad y = -1, \quad z = 3 \]
Substituting \( r = \frac{15}{11} \):
\[ x = 3\left(\frac{15}{11}\right) - 2 = \frac{45}{11} - \frac{22}{11} = \frac{23}{11} \]
\[ y = 2\left(\frac{15}{11}\right) - 1 = \frac{30}{11} - \frac{11}{11} = \frac{19}{11} \]
\[ z = 2\left(\frac{15}{11}\right) + 3 = \frac{30}{11} + \frac{33}{11} = \frac{63}{11} \]

Now we check if these points lie on the second line:
\[ \frac{x + 7}{5} = \frac{y + 5}{4} = z - 2 \]
For \( r = 0 \):
\[ \frac{-2 + 7}{5} = 1 \quad \text{and} \quad \frac{-1 + 5}{4} = 1 \quad \text{and} \quad 3 - 2 = 1 \]
Thus, the point \( (-2, -1, 3) \) lies on the second line.

### Conclusion for Statement 1:
The first statement is **true**.

### Statement 2:
We need to find the distance \( d \) between the point \( (-1, -5, -10) \) and the point of intersection of the line:
\[ \frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12} \]
with the plane \( x - y + z = 5 \).

Let \( r \) be the common ratio:
\[ x = 3r + 2, \quad y = 4r - 1, \quad z = 12r + 2 \]
Substituting these into the plane equation:
\[ (3r + 2) - (4r - 1) + (12r + 2) = 5 \]
Simplifying:
\[ 3r + 2 - 4r + 1 + 12r + 2 = 5 \]
\[ 11r + 5 = 5 \implies 11r = 0 \implies r = 0 \]
Thus, the point of intersection is:
\[ x = 2, \quad y = -1, \quad z = 2 \]

Now we calculate the distance \( d \):
\[ d = \sqrt{(2 - (-1))^2 + (-1 - (-5))^2 + (2 - (-10))^2} \]
Calculating each term:
1. \( (2 + 1)^2 = 3^2 = 9 \)
2. \( (-1 + 5)^2 = 4^2 = 16 \)
3. \( (2 + 10)^2 = 12^2 = 144 \)

Combining these:
\[ d^2 = 9 + 16 + 144 = 169 \]
Thus, \( d = \sqrt{169} = 13 \).

### Conclusion for Statement 2:
The second statement is also **true**.

### Final Conclusion:
Both statements are true, but statement 2 does not explain statement 1.