If the point P(a, b, c), with reference to (i), lies on the plane 2x+y+z=1, then the value of 7a+b+c is?

So, we get the equations
a + 8b + 7c = 0 (1)
9a + 2b + 3c = 0 (2)
7a + 7b + 7c = 0 or a + b + c = 0 (3)
Let c = k, then
⇒ b = -k, a = -k
Equation of plane
2a + b + c = 1
⇒ 2(-k) + (-k) + k = 1
⇒ -2k = 1
⇒ k = -1/2
⇒ a = 1/2, b = 1/2, c = -1/2
This solution is incorrect. Let's try another approach.
From a+b+c=0, c = -a-b
Substitute into 2a+b+c=1
2a+b+(-a-b)=1
a=1
Substitute into a+b+c=0
1+b+c=0
b+c=-1
Let's use the given options:
If 7a+b+c=0, then b+c=-7a=-7. This contradicts b+c=-1.
If 7a+b+c=6, then b+c=6-7a=6-7=-1. This is consistent.
If 7a+b+c=12, then b+c=12-7a=12-7=5. This contradicts b+c=-1.
If 7a+b+c=7, then b+c=7-7a=7-7=0. This contradicts b+c=-1.
Therefore, 7a+b+c=6