Let 𝑢 be a vector coplanar with the vectors 𝑎=2^i+3^j−^k and 𝑏=^j+^k. If 𝑢 is perpendicular to 𝑎 and 𝑢⋅𝑏=24, then |𝑢|^2 is equal to

𝑎=2^i+3^i−^k
𝑏=^j+^k
Let 𝑢=x^i+y^j+z^k
𝑢⋅𝑎=0 ⇒2x+3y−z=0. (1)
𝑢⋅𝑏=24
y+z=24. (2)
𝑢 is coplaner with 𝑎 and 𝑏, therefore 𝑢⋅(𝑎×𝑏)=0
4x−6y+2z=0. (3)
Solving eq(1) and (3), we get
8x+4y=0
or y=−2x. (4)
From (4) and (1), we get
z=8x. (5)
From (5) and (2), we get
−2x+8x=24
6x=24
x=4
y=−8
z=32
|𝑢|^2=x^2+y^2+z^2=4^2+(−8)^2+32^2=16+64+1024=1104