If the lines `(x-1)/(2)=(y+2)/(a)=(z-3)/(10)and (x-2)/(3)=(y+3)/(-6)=(z+4)/6` are parallel to each other, then

Options

A. a-2,b=5
B. a=4,b=5
C. a=-4,b=15
D. a=4,b=15

Explanation

To determine the value of \(a\) for which the lines

\[ \frac{x-1}{2} = \frac{y+2}{a} = \frac{z-3}{10} \]

and

\[ \frac{x-2}{3} = \frac{y+3}{-6} = \frac{z+4}{6} \]

are parallel, we need to find the direction ratios of both lines and set them proportional to each other.

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

For the first line, the direction ratios can be extracted from the coefficients:

- \(a_1 = 2\)
- \(b_1 = a\)
- \(c_1 = 10\)

For the second line, the direction ratios are:

- \(a_2 = 3\)
- \(b_2 = -6\)
- \(c_2 = 6\)

### Step 2: Set up the proportionality condition

Since the lines are parallel, the direction ratios must be proportional. Therefore, we can write:

\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]

Substituting the values we have:

\[ \frac{2}{3} = \frac{a}{-6} = \frac{10}{6} \]

### Step 3: Solve for \(a\)

From the equation \(\frac{2}{3} = \frac{a}{-6}\):

Cross-multiplying gives:

\[ 2 \cdot (-6) = 3a \implies -12 = 3a \implies a = -4 \]

### Step 4: Solve for \(b\)

From the equation \(\frac{2}{3} = \frac{10}{6}\):

Cross-multiplying gives:

\[ 2 \cdot 6 = 3 \cdot 10 \implies 12 = 30 \]

This is not needed for our solution since we are only asked for \(a\).

### Final Answer

Thus, the value of \(a\) for which the lines are parallel is:

\[ \boxed{-4} \]