Given α = 3i + 4j + 5k and β = 2i + j + 4k

Any vector parallel to α is  β₁ = aα where a is a constant.

Let β₂ = pi + qj + rk is perpendicular to α.

Then β₂ * α = 0

=> (pi + qj + rk)*(3i + 4j + 5k) = 0

=> 3p + 4q + 5r = 0 ..............1

Now, β = β₁ + β₂ 

=> 2i + j + 4k = aα + (pi + qj + rk) 

=> 2i + j + 4k = a(3i + 4j + 5k) + (pi + qj + rk)

=> 2i + j + 4k = i(3a + p) + j(4a + q) + k(5a + r)

So,  3a + p = 2 ............2

      4a + q = 1 ............3

and 5a + r = 4 .............4

From 2, p = 2 - 3a

From 3, q = 1 - 4a

From 4, r = 4 - 5a

Pt value of p, q and r in equation 1, we get

      3(2- 3a) + 4(1 - 4a) + 5(4 - 5a) = 0

=> 6 - 9a + 4 - 16a + 20 - 25a = 0

=> 30 - 50a = 0

=> 50a = 30

=> a = 30/50

=> a = 3/5

Now, 

p = 2 - 3a = 2 - 3*3/5 = 2 - 9/5 = 1/5

q = 1 - 4a = 1 - 4*3/5 = 1 - 12/5 = -7/5

r = 4 - 5a = 4 - 5*3/5 = 4 - 3 = 1

Now, β = aα + β₁ + β₂

=> β = a((3i + 4j + 5k)) + i(3a + p) + j(4a + q) + k(5a + r)

=> β = {3(3i + 4j + 5k)/5} + {i(3*3/5 + 1/5) + j(4*3/5 - 7/5) + k(5*3/5 + 1)}

=> 2i + j + 4k = {9i/5 + 12i/5 + 3k} + {i(9/5 + 1/5) + j(12/5 - 7/5) + k(3 + 1)}

=> 2i + j + 4k = {9i/5 + 12i/5 + 3k} + {2i + j + 4k}