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(B) Let the two lines be $L_1: x = y + a = z = \lambda$ and $L_2: x + a = 2y = 2z = 2\mu$.From $L_1$,any point $P$ is $(\lambda, \lambda - a, \lambda)

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

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(B) Let the two lines be $L_1: x = y + a = z = \lambda$ and $L_2: x + a = 2y = 2z = 2\mu$.From $L_1$,any point $P$ is $(\lambda, \lambda - a, \lambda)$.From $L_2$,any point $Q$ is $(2\mu - a, \mu, \mu)$.The direction ratios of the line $PQ$ are $(2\mu - a - \lambda, \mu - \lambda + a, \mu - \lambda)$.Since the line $PQ$ is proportional to $2, 1, 2$,we have:$\frac{2\mu - a - \lambda}{2} = \frac{\mu - \lambda + a}{1} = \frac{\mu - \lambda}{2} = k$.From $\frac{\mu - \lambda}{2} = k$,we get $\mu - \lambda = 2k$.Substituting this into the second ratio: $2k + a = k \implies k = -a$.Then $\mu - \lambda = -2a$.From the first ratio: $\frac{2\mu - \lambda - a}{2} = -a \implies 2\mu - \lambda = a$.Solving $\mu - \lambda = -2a$ and $2\mu - \lambda = a$,we get $\mu = 3a$ and $\lambda = 3a$.For $P$,substituting $\lambda = 3a$: $P = (3a, 3a - a, 3a) = (3a, 2a, 3a)$.For $Q$,substituting $\mu = a$ (Wait,re-solving: $2\mu - \lambda = a$ and $\mu - \lambda = -2a \implies \mu = 3a, \lambda = 5a$ is incorrect. Let's re-evaluate: $\frac{2\mu - a - \lambda}{2} = \frac{\mu - \lambda + a}{1} = \frac{\mu - \lambda}{2}$.Let $\mu - \lambda = m$. Then $\frac{2\mu - a - \lambda}{2} = \frac{m+a}{1} = \frac{m}{2}$.$m = 2m + 2a \implies m = -2a$.$2\mu - a - \lambda = 2(m+a) = 2(-2a+a) = -2a \implies 2\mu - \lambda = -a$.With $\mu - \lambda = -2a$ and $2\mu - \lambda = -a$,we get $\mu = a$ and $\lambda = 3a$.Thus,$P = (3a, 2a, 3a)$ and $Q = (2(a) - a, a, a) = (a, a, a)$.

$A$ line with direction ratios proportional to $2, 1, 2$ meets each of the lines $x = y + a = z$ and $x + a = 2y = 2z$. The coordinates of each of the points of intersection are given by

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