If the lines $\frac{x-1}{k}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x-2}{3}=\frac{y-3}{k}=\frac{z-1}{2}$ intersect in a point, then the integral value of k is:

Question

Accepted Answer

To determine the integral value of $ k $ for which the given lines intersect, we need to analyze the parametric equations of the lines.

The first line is given by:
$$
\frac{x-1}{k} = \frac{y-2}{2} = \frac{z-3}{3} = t
$$
This can be rewritten in parametric form as:
$$
x = kt + 1, \quad y = 2t + 2, \quad z = 3t + 3
$$

The second line is given by:
$$
\frac{x-2}{3} = \frac{y-3}{k} = \frac{z-1}{2} = s
$$
This can be rewritten in parametric form as:
$$
x = 3s + 2, \quad y = ks + 3, \quad z = 2s + 1
$$

For the lines to intersect, there must exist values of $ t $ and $ s $ such that the coordinates $ (x, y, z) $ are the same for both lines. Therefore, we equate the parametric equations:

1. $ kt + 1 = 3s + 2 $
2. $ 2t + 2 = ks + 3 $
3. $ 3t + 3 = 2s + 1 $

From equation 3:
$$
3t + 3 = 2s + 1 \implies 3t = 2s - 2 \implies t = \frac{2s - 2}{3}
$$

Substitute $ t = \frac{2s - 2}{3} $ into equation 1:
$$
k\left(\frac{2s - 2}{3}\right) + 1 = 3s + 2
$$
$$
\frac{k(2s - 2)}{3} + 1 = 3s + 2
$$
$$
k(2s - 2) + 3 = 9s + 6
$$
$$
2ks - 2k + 3 = 9s + 6
$$
$$
2ks - 9s = 2k + 3
$$
$$
s(2k - 9) = 2k + 3
$$
$$
s = \frac{2k + 3}{2k - 9}
$$

Substitute $ t = \frac{2s - 2}{3} $ into equation 2:
$$
2\left(\frac{2s - 2}{3}\right) + 2 = ks + 3
$$
$$
\frac{4s - 4}{3} + 2 = ks + 3
$$
$$
4s - 4 + 6 = 3ks + 9
$$
$$
4s + 2 = 3ks + 9
$$
$$
4s - 3ks = 7
$$
$$
s(4 - 3k) = 7
$$
$$
s = \frac{7}{4 - 3k}
$$

Equating the two expressions for $ s $:
$$
\frac{2k + 3}{2k - 9} = \frac{7}{4 - 3k}
$$

Cross-multiply:
$$
(2k + 3)(4 - 3k) = 7(2k - 9)
$$
$$
8k + 12 - 6k^2 - 9k = 14k - 63
$$
$$
-6k^2 - k + 12 = 14k - 63
$$
$$
-6k^2 - 15k + 75 = 0
$$
$$
6k^2 + 15k - 75 = 0
$$

Divide the entire equation by 3:
$$
2k^2 + 5k - 25 = 0
$$

Using the quadratic formula $ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $, where $ a = 2 $, $ b = 5 $, $ c = -25 $:
$$
k = \frac{-5 \pm \sqrt{5^2 - 4 \times 2 \times (-25)}}{2 \times 2}
$$
$$
k = \frac{-5 \pm \sqrt{25 + 200}}{4}
$$
$$
k = \frac{-5 \pm \sqrt{225}}{4}
$$
$$
k = \frac{-5 \pm 15}{4}
$$

The solutions are:
$$
k = \frac{10}{4} = 2.5 \quad \text{and} \quad k = \frac{-20}{4} = -5
$$

Since we are looking for an integral value of $ k $, the correct answer is $ k = -5 $.

Thus, Option 2 is correct. The other options do not satisfy the condition for the lines to intersect at an integral value of $ k $.