Given the system of linear equations.
8x+y+4z=−2
x+y+z=0
λx−3y=μ
has infinitely many solutions, so
D=∣81λ11−3410∣=0⇒λ=4
Also D1=D2=D3=0
D1=∣−20μ11−3410∣=0⇒−6−3μ=0
So μ=−2
Hence the point will be (4,−2,−21)
Now distance of point (4,−2,−21) from plane 8x+y+4z+2=0
=∣82+12+4232−2−2+2∣=∣930∣=310