Someone please help me…vectors

Let L2 be the line passes through the point Q(10,1,6),has a point intersection with L1,and orthogonal to L1.Find the vector equation of L2.

L1: x= 1+4t y=5-4t z=-1+5t

Let be a direction vector for L2.

So, the parametric equations for L2 are
x = 10 + as, y = 1 + bs, z = 6 + cs
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Since L1 has direction vector <4, -4, 5>, and L1 is orthogonal to L2, we have
4a - 4b + 5c = 0.

Next, since L1 and L2 intersect, we have
10 + as = 1 + 4t, 1 + bs = 5 - 4t, and 6 + cs = -1 + 5t
==> as = -9 + 4t, bs = 4 - 4t, and cs = -7 + 5t

Substitute this into 4a - 4b + 5c = 0 <==> 4as - 4bs + 5cs = 0:
4(-9 + 4t) - 4(4 - 4t) + 5(-7 + 5t) = 0
==> -87 + 57t = 0
==> t = 87/57 = 29/19.

So, as = -9 + 4 * 29/19, bs = 4 - 4 * 29/19, and cs = -7 + 5 * 29/19
==> as = -55/19, bs = -40/19, and cs = 12/19
==> = <-55, -40, 12> (or any scalar multiple thereof)

Hence, the vector equation of L2 is
L2(s) = <10 - 55s, 1 - 40s, 6 + 12s>.

I hope this helps!