If the system of equations
$$x + 5y + 6z = 4$$,
$$2x + 3y + 4z = 7$$,
$$x + 6y + az = b$$
has infinitely many solutions, then the point $$(a, b)$$ lies on the line :

The three simultaneous equations are
$$x + 5y + 6z = 4 \qquad -(1)$$
$$2x + 3y + 4z = 7 \qquad -(2)$$
$$x + 6y + az = b \qquad\;\; -(3)$$

For infinitely many solutions we need
$$\text{rank}(\mathbf A)=\text{rank}(\mathbf A| \mathbf B)\lt 3,$$
where $$\mathbf A$$ is the coefficient matrix. Hence $$\det(\mathbf A)=0.$

The coefficient matrix and its determinant:

$$\mathbf A=$$\begin{vmatrix}$$ 1 & 5 & 6\\ 2 & 3 & 4\\ 1 & 6 & a \end{vmatrix},\qquad \Delta=$$\begin{vmatrix}$$ 1 & 5 & 6\\ 2 & 3 & 4\\ 1 & 6 & a \end{vmatrix}.$$

Expanding along the first row,

$$\Delta =1$$\begin{vmatrix}$$3 & 4\\ 6 & a\end{vmatrix} -5$$\begin{vmatrix}$$2 & 4\\ 1 & a\end{vmatrix} +6$$\begin{vmatrix}$$2 & 3\\ 1 & 6\end{vmatrix}$$ $$=(3a-24)-5(2a-4)+6(12-3)$$ $$=(3a-24)-(10a-20)+54$$ $$=-7a+50.$$

Setting $$\Delta=0$$ gives

$$-7a+50=0\Longrightarrow a=$$\frac{50}{7}$$.$$

Because $$\det(\mathbf A)=0,$$ the three rows of $$\mathbf A$$ are linearly dependent. Let the third row be a linear combination of the first two:

$$$$\text{Row}_3=\lambda\$$,$$\text{Row}_1+\mu\$$,$$\text{Row}_2$$.$$

Comparing the coefficients of $$x$$ and $$y$$:

$$$$\lambda$$+2$$\mu$$=1 \qquad -(4)$$ $$5$$\lambda$$+3$$\mu$$=6 \qquad -(5)$$

Solving (4) and (5),

From (4): $$$$\lambda$$=1-2$$\mu$$.$$ Substitute in (5): $$5(1-2$$\mu$$)+3$$\mu$$=6\; \Longrightarrow\; -7$$\mu$$=1\; \Longrightarrow\; $$\mu$$=-$$\frac{1}{7}$$.$$ Hence $$$$\lambda$$=1-$$\frac{-2}{7}=\frac{9}{7}$$.$$

Now equate the $$z$$-coefficients and the constants.

For $$z$$: $$6$$\lambda$$+4$$\mu$$=a \;\Longrightarrow\;6\!$$\left$$($$\frac{9}{7}$$$$\right$$)+4\!$$\left$$(-$$\frac{1}{7}$$$$\right$$)=$$\frac{50}{7}$$,$$
which is exactly the value already obtained for $$a$$, confirming consistency.

For the constants: $$4$$\lambda$$+7$$\mu$$=b$$

$$b=4\!$$\left$$($$\frac{9}{7}$$$$\right$$)+7\!$$\left$$(-$$\frac{1}{7}$$$$\right$$)=$$\frac{36}{7}$$-1=$$\frac{29}{7}$$.$$

Thus the required point is

$$(a,b)=$$\left$$($$\frac{50}{7}$$,\;$$\frac{29}{7}$$$$\right$$).$$

Check which of the given straight lines contains this point:

For $$x-y=3: \; $$\frac{50}{7}-\frac{29}{7}$$=3 \quad $$\text{(satisfied)}$$.$$
For $$y-x=3: \; $$\frac{29}{7}-\frac{50}{7}$$=-3 \quad $$\text{(not satisfied)}$$.$$
For $$x+y=11: \; $$\frac{50}{7}+\frac{29}{7}=\frac{79}{7}$$$$\approx$$11.29 $$\neq$$11.$$
For $$x+y=12: \; $$\frac{79}{7}$$$$\approx$$11.29 $$\neq$$12.$$

The only line through the point $$(a,b)$$ is

$$x-y=3.$$

Hence, the correct choice is
Option B which is: $$x - y = 3$$.