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(B) The given lines are $L_1: x = y + a = z$ and $L_2: x + a = 2y = 2z$.Rewrite $L_1$ as $\frac{x}{1} = \frac{y+a}{1} = \frac{z}{1} = k_1$. Thus,any p

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$(3a, 2a, 3a), (a, a, a)$

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(B) The given lines are $L_1: x = y + a = z$ and $L_2: x + a = 2y = 2z$.Rewrite $L_1$ as $\frac{x}{1} = \frac{y+a}{1} = \frac{z}{1} = k_1$. Thus,any point $P$ on $L_1$ is $(k_1, k_1 - a, k_1)$.Rewrite $L_2$ as $\frac{x+a}{2} = \frac{y}{1} = \frac{z}{1} = k_2$. Thus,any point $Q$ on $L_2$ is $(2k_2 - a, k_2, k_2)$.The direction ratios of the line $PQ$ are $(2k_2 - a - k_1, k_2 - (k_1 - a), k_2 - k_1)$,which simplifies to $(2k_2 - a - k_1, k_2 - k_1 + a, k_2 - k_1)$.Given the direction ratios are in the ratio $2 : 1 : 2$,we have:$\frac{2k_2 - a - k_1}{2} = \frac{k_2 - k_1 + a}{1} = \frac{k_2 - k_1}{2}$.From $\frac{k_2 - k_1 + a}{1} = \frac{k_2 - k_1}{2}$,we get $2k_2 - 2k_1 + 2a = k_2 - k_1$,which implies $k_2 - k_1 = -2a$,or $k_1 - k_2 = 2a$.From $\frac{2k_2 - a - k_1}{2} = \frac{k_2 - k_1}{2}$,we get $2k_2 - a - k_1 = k_2 - k_1$,which implies $k_2 = a$.Substituting $k_2 = a$ into $k_1 - k_2 = 2a$,we get $k_1 = 3a$.Thus,the points of intersection are $P = (3a, 3a - a, 3a) = (3a, 2a, 3a)$ and $Q = (2a - a, a, a) = (a, a, a)$.

If the direction ratios of a line are in the ratio $2 : 1 : 2$ and it intersects the lines $x = y + a = z$ and $x + a = 2y = 2z$,then the coordinates of the points of intersection are:

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