The equation $x + 2y + 3z = 1,x - y + 4z = 0$ and $2x + y + 7z = 1$ have
A) One solution only
B) Two solutions only
C) No solution
D) Infinitely many solutions

Hint: Express the equations in matrix form and then try to reduce the equations in the echelon form using some row operations and then analyze the obtained matrix to get the desired result.

Complete step-by-step solution:
Consider the given equations:
$x + 2y + 3z = 1$
$x - y + 4z = 0$
$2x + y + 7z = 1$
The goal of the problem is to find that the given system has only one solution or only two solutions or no solution or infinitely many solutions.
Express the given equation in form of augmented matrix with the coefficient of $x$ in the first column, coefficient of $y$ in the second column, coefficient of $z$ in the third column, and the constant terms on the fourth column.
Then the augmented matrix is given as:
$\left[ {\begin{array}{*{20}{c}}
  1&2&3&1 \\
  1&{ - 1}&4&0 \\
  2&1&7&1
\end{array}} \right]$
Now, try to reduce the matrix in echelon form having some row operations.
Apply the row operations,${R_2} \to {R_2} - {R_1}{\text{ and }}{{\text{R}}_3} \to {R_3} - 2{R_1}$, then the above matrix is given as:
$\left[ {\begin{array}{*{20}{c}}
  1&2&3&1 \\
  0&{ - 3}&1&{ - 1} \\
  0&{ - 3}&1&{ - 1}
\end{array}} \right]$
Now, apply the row operation,${R_2} \to {R_2} - {R_3}$, now the obtained matrix is given as:
$\left[ {\begin{array}{*{20}{c}}
  1&2&3&1 \\
  0&0&0&0 \\
  0&{ - 3}&1&{ - 1}
\end{array}} \right]$
Now, express the matrix again in the form of an equation. So, the formed equations from the above matrix are:
$x + 2y + 3z = 1{\text{ and }} - 3y + z = - 1$
Where $z$ can be any real number, so the$z$is the parameter of the above equations. Different real values of $z$ give a different solution to the system. It means that real values of $z$gives the infinitely many solutions of the system.
Therefore, there are infinitely many solutions to the given system. Thus, we can say that the equations:
$x + 2y + 3z = 1,x - y + 4z = 0$ and $2x + y + 7z = 1$ have infinitely many solutions.

Hence, the option (D) is the correct option.

Note:
While reducing the matrix in echelon form, we have to notice that the first leading coefficient is always to the right of the first nonzero number in the row above and the rows consisting of all zero entries are at the bottom of the matrix.