You need to multiply the vectors `(0,-1,3),(1,1,1),(1,2,5)` by scalars a,b,c and then you need to perform addition of first two vectors and then you need to subtract the third vector such that:

`a*0*bari + a*(-1)*barj + a*3*bark + b*1*bari + b*1*barj + b*1*bar k- c*1*bari - c*2*bar k- c*5*bar k = -2*bar i + 3*bar j - 8*bar k`

`-a*barj + 3a*bark + b*bari + b*barj + b*bark - c*bari - 2c*barj - 5c*bark = -2*bar i + 3*bar j - 8*bar k`

You need to factor out `bari, barj, bark` , to the left such that:

`(b-c)bari + (-a+b-2c)barj + (3a+b-5c)bark = -2*bar i + 3*bar j - 8*bar k`

Equating coefficients of like unit vectors yields:

`b - c = -2 =gt b = c-2`

`-a+b-2c = 3`

`3a + b - 5c = -8`

You need to multiply the second equation by 3 such that:

`-3a + 3b - 6c = 9`

You need to add this equation to `3a + b - 5c = -8` such that:

`3a + b - 5c - 3a + 3b - 6c = 9 -8`

`4b - 11c = -1`

You need to multiply the equation `b-c=-2` by -4 such that:

`-4b + 4c = 8`

You need to add this equation to `4b - 11c = -1` such that:

`4b - 11c +4c - 4b= 8-1`

`-7c = 7 =gt c = -1`

You need to substitute -1 for c in the equation `b - c = -2 ` such that:

`b + 1 = -2 =gt b = -2 -1 =gt b = -3`

You need to substitute -3 for b and -1 for c in `-a+b-2c = 3` such that:

`-a - 3 + 2 = 3 =gt -a = 4 =gt a = -4`

**Hence, evaluating scalars `a,b,c` yields `a=-4, b=-3, c=-1` .**

It is given that a(0,-1,3) + b(1,1,1) - c(1,2,5) = (-2,3,-8). This gives the set of linear equations:

b - c = -2 ...(1)

-a + b - 2c = 3 ...(2)

3a + b - 5c = -8 ...(3)

b - c = -2 => b = c - 2

Substitute in (2) and (3)

=> -a + c - 2 - 2c = 3 and 3a + c - 2 - 5c = -8

=> -a - c = 5 and 3a - 4c = -6

Substitute a = -5 - c in 3a - 4c = -6

=> -15 - 3c - 4c = -6

=> -7c = 9

=> c = -9/7

a = -5 + 9/7 = -26/7

b = -9/7 - 2 = -23/7

**The value of a = `-26/7` , b =` -23/7` and c =` -9/7` **