A line passes through (2, -1, 3) and it is perpendicular to the lines r = (i + j - k) + λ(2i - j + k) and r = (2i - j - k) + μ(i + 2j + 2k). Obtain its equation in vector and Cartesian form.

Line L is passing through point = (2i - j + 3k)
If L1 → r = (i + j - k) + λ(2i - j + k)
L2 → r = (2i - j - k) + μ(i + 2j + 2k)
Given that line is perpendicular to L1 and L2
Let the line L = (a1, a2, a3)
The equation of L in vector form → r = (2i - j - k) + p(a1i + a2j + a3k) p is any constant.
So by condition that L is perpendicular to L1
2a1 - a2 + a3 = 0 — (1)
L ⊥ L2
So, a1 + 2a2 + 2a3 = 0 — (2)
Solve (1) and (2)
3a1 + 3a3 = 0 → a3 = -a1
Put it in (2)
a1 + 2a2 - a1 = 0
a2 = 0
So, L = (a1, 0, -a1)
So we can say DR of L = (1, 0, -1)
So equation of L in vector form:
r = (2i - j + 3k) + k(i - k)
Cartesian form is x - 1 = y + 1 = z - 3/-1