If the system of linear equations
$$2x+ y+7z = a$$
$$6x-2y+11z = b$$
$$2x-y+3z = c$$ 
has infinite number of solutions, then $$a, b, c$$ must satisfy

Let's represent the system of linear equations
$$2x+ y+7z = a$$
$$6x-2y+11z = b$$
$$2x-y+3z = c$$ in the Matrix Equation form

$$\begin{bmatrix}2 & 1 &7 & a  \\6 & -2 & 11 & b &\\2 & -1 &3 & c \end{bmatrix}$$

Order of the matrix = 3

Now represent the matrix in echelon form 

$$\begin{bmatrix}2 & 1 &7 & a  \\0 & -5 & -10 & b-3a &\\0 & -2 &-4 & c-a \end{bmatrix}$$

$$\begin{bmatrix}2 & 1 &7 & a  \\6 & -2 & 11 & b &\\0 & 0 &0 & 5c-5a-2b+6a \end{bmatrix}$$

Rank of the matrix = No of Non-zero rows after transforming the matrix into echelon form

The system of linear equations will have infinite solutions only if Rank of the matrix is not equal to Order of the Matrix

$$\therefore$$ 5c-5a-2b+6a=0

a-2b+5c=0

D is the correct answer.