Determine whether the lines L1:x=23+6s,y=17+5s,z=-2+s,and L2:x=-9+7t,y=-12+7t,z=-13+4t are intersect,are skew or are parallel.If they intersect,determine the point of intersection;if not leave the remaining answer blanks empty.

To determine whether the lines L1 and L2 intersect, are skew, or are parallel, we need to check if they have a common point (intersect), have no common points but are not parallel (skew), or have no common points and are parallel (parallel).

The parametric equations of the lines are:

L1: x = 23 + 6s, y = 17 + 5s, z = -2 + s L2: x = -9 + 7t, y = -12 + 7t, z = -13 + 4t

We can set the x, y, and z components equal to each other and solve for s and t:

23 + 6s = -9 + 7t 17 + 5s = -12 + 7t -2 + s = -13 + 4t

Solving these equations, we get:

s = (7t - 32) / 6 s = (7t - 29) / 5 s = 4t - 11

Setting these equal to each other, we get:

(7t - 32) / 6 = (7t - 29) / 5 (7t - 32) / 6 = 4t - 11

Solving these equations, we get:

t = 18, s = 1

Substituting these values back into the equations for L1 and L2, we get:

L1: x = 23 + 6(1) = 29, y = 17 + 5(1) = 22, z = -2 + 1 = -1 L2: x = -9 + 7(18) = 117, y = -12 + 7(18) = 114, z = -13 + 4(18) = 59

Since the points (29, 22, -1) and (117, 114, 59) are not the same, the lines L1 and L2 do not intersect.

Next, we check if the lines are parallel by comparing their direction vectors. The direction vector of L1 is (6, 5, 1) and of L2 is (7, 7, 4). Since these vectors are not proportional, the lines are not parallel.

Therefore, the lines L1 and L2 are skew.