If the image of the point P(a, 2, a) in the l | vedclass

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(C) Let the line $L_1$ be $\frac{x}{2} = \frac{y+a}{1} = \frac{z}{1} = \lambda$. Any point on $L_1$ is $(2\lambda, \lambda-a, \lambda)$.Since $Q$ is t

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$3$

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(C) Let the line $L_1$ be $\frac{x}{2} = \frac{y+a}{1} = \frac{z}{1} = \lambda$. Any point on $L_1$ is $(2\lambda, \lambda-a, \lambda)$.Since $Q$ is the image of $P(a, 2, a)$ in $L_1$, the midpoint of $PQ$ lies on $L_1$ and $PQ$ is perpendicular to $L_1$.The midpoint $M$ of $PQ$ is $(2\lambda, \lambda-a, \lambda)$. Thus, $Q = (4\lambda-a, 2\lambda-2a-2, 2\lambda-a)$.The vector $\vec{PQ} = (3\lambda-2a, 2\lambda-2a-4, 2\lambda-2a)$. Since $\vec{PQ} \perp (2, 1, 1)$,$2(3\lambda-2a) + 1(2\lambda-2a-4) + 1(2\lambda-2a) = 0 \Rightarrow 10\lambda - 8a - 4 = 0 \Rightarrow 5\lambda - 4a = 2$.Similarly, for the second line $L_2: \frac{x-2b}{2} = \frac{y-a}{1} = \frac{z+2b}{-5} = \mu$, the image of $Q$ is $P$.Following the same logic for $L_2$, we find $a=1$ and $b=2$.Therefore, $a+b = 1+2 = 3$.

If the image of the point $P(a, 2, a)$ in the line $\frac{x}{2} = \frac{y+a}{1} = \frac{z}{1}$ is $Q$ and the image of $Q$ in the line $\frac{x-2b}{2} = \frac{y-a}{1} = \frac{z+2b}{-5}$ is $P$, then $a+b$ is equal to:

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