Let p,q,r in `R` satisfies `[p q r][(1,8,7),(9,2,3),(7,7,7)] = [0 0 0] .....(i)`
`{:(,"List-I",,"List-I"),((P),"If the point M(p.q.r) with reference to (i) lies on the curve" 2x+y+z=1 then (7p+q+r) "is equal to",(1),-2),((Q),"Let" omega(ne1) "cube root of unity with" lm(omega) gt0.If p=2 "with q and r satisfying" (i) then (3/(omega^(p))+1/(omega^(q))+3/(omega^(r))) "is equal to ",(2),7),((R),"Let q=6 with p and r satisfying"(i). if alpha and beta "are roots of quadratic equation " px^(2)+qx+r=0 " then" Sigma_(n=0)^(oo) (1/(alpha)+1/(beta))^(n) " is equal to ",(3),6):}`

Explanation

To solve the problem, we need to analyze the given equation and the conditions provided.

### Step 1: Analyze the given matrix equation
We have the equation:
\[ [p \quad q \quad r] \begin{pmatrix} 1 & 8 & 7 \\ 9 & 2 & 3 \\ 7 & 7 & 7 \end{pmatrix} = [0 \quad 0 \quad 0] \]
This implies that the linear combination of the rows of the matrix with coefficients \(p\), \(q\), and \(r\) results in the zero vector.

### Step 2: Set up the system of equations
This can be expressed as a system of equations:
1. \( p + 8q + 7r = 0 \)
2. \( 9p + 2q + 3r = 0 \)
3. \( 7p + 7q + 7r = 0 \)

### Step 3: Simplify the third equation
From the third equation, we can simplify it:
\[ 7p + 7q + 7r = 0 \implies p + q + r = 0 \quad \text{(divide by 7)} \]
This gives us:
\[ r = -p - q \]

### Step 4: Substitute \(r\) into the first two equations
Now, substitute \(r = -p - q\) into the first two equations.

1. Substitute into the first equation:
\[ p + 8q + 7(-p - q) = 0 \implies p + 8q - 7p - 7q = 0 \implies -6p + q = 0 \implies q = 6p \]

2. Substitute into the second equation:
\[ 9p + 2q + 3(-p - q) = 0 \implies 9p + 2q - 3p - 3q = 0 \implies 6p - q = 0 \implies q = 6p \]

### Step 5: Find \(r\) in terms of \(p\)
Now, we have \(q = 6p\). Substitute this into \(r = -p - q\):
\[ r = -p - 6p = -7p \]

### Step 6: Check the conditions for the point \(M(p, q, r)\)
Now we have:
\[ M(p, q, r) = (p, 6p, -7p) \]
We need to check if this point lies on the curve \(2x + y + z = 1\):
\[ 2(p) + 6p - 7p = 1 \implies 2p - p = 1 \implies p = 1 \]
Thus, substituting \(p = 1\):
\[ q = 6(1) = 6, \quad r = -7(1) = -7 \]

### Step 7: Calculate \(7p + q + r\)
Now we can find \(7p + q + r\):
\[ 7(1) + 6 + (-7) = 7 + 6 - 7 = 6 \]

### Conclusion
The value of \(7p + q + r\) is \(6\).