For the pair of equations $\lambda x + 3y = -7$ and $2x + 6y = 14$ to have infinitely many solutions,the value of $\lambda$ should be $1$. Is the statement true? Give reasons.

(B) The given pair of linear equations is $\lambda x + 3y + 7 = 0$ and $2x + 6y - 14 = 0$.
For a system of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to have infinitely many solutions,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Here,$a_1 = \lambda, b_1 = 3, c_1 = 7$ and $a_2 = 2, b_2 = 6, c_2 = -14$.
Substituting these values into the condition,we get $\frac{\lambda}{2} = \frac{3}{6} = \frac{7}{-14}$.
Simplifying the ratios,we get $\frac{\lambda}{2} = \frac{1}{2} = -\frac{1}{2}$.
Since $\frac{1}{2} \neq -\frac{1}{2}$,the condition $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ can never be satisfied for any value of $\lambda$.
Therefore,the statement is false.