Let →c=x→i+y→j+z→k
→a×→c=∣∣
∣
∣∣→i→j→k111xyz∣∣
∣
∣∣
=→i(z−y)−→j(z−x)+→k(y−x)
Now, →a×→c=→b
Given →b=→j−→k
Therefore, z - y = 0, x -z = 1, y - x = -1
y = z and x - y = 1----- (1)
Again, →a.→c=3
x+y+z=3
x+y+y=3
x+2y=3----(2)
Solving 1 and 2,
x + 2y = 3
x - y = 1
__________
2y=2
y=23=z
Therefore, x=1+y=53
So, →c=53→i+23→j+23→k