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(B) The equation of the plane is $2x - y + z = b$.Let $P = (1, 3, a)$ and $Q = (-3, 5, 2)$. The midpoint $R$ of the line segment $PQ$ lies on the plan

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$1$

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(B) The equation of the plane is $2x - y + z = b$.Let $P = (1, 3, a)$ and $Q = (-3, 5, 2)$. The midpoint $R$ of the line segment $PQ$ lies on the plane.$R = \left( \frac{1 - 3}{2}, \frac{3 + 5}{2}, \frac{a + 2}{2} \right) = (-1, 4, \frac{a + 2}{2})$.Since $R$ lies on the plane $2x - y + z = b$,we have:$2(-1) - 4 + \frac{a + 2}{2} = b$$-6 + \frac{a + 2}{2} = b \Rightarrow a + 2 = 2b + 12 \Rightarrow a = 2b + 10 \quad \dots(i)$Also,the vector $\vec{PQ} = (-3 - 1, 5 - 3, 2 - a) = (-4, 2, 2 - a)$ is parallel to the normal vector of the plane $\vec{n} = (2, -1, 1)$.Therefore,$\frac{-4}{2} = \frac{2}{-1} = \frac{2 - a}{1}$.$-2 = -2 = 2 - a \Rightarrow a = 4$.Substituting $a = 4$ into equation $(i)$:$4 = 2b + 10 \Rightarrow 2b = -6 \Rightarrow b = -3$.Thus,$|a + b| = |4 + (-3)| = |1| = 1$.

Let the mirror image of the point $P(1, 3, a)$ with respect to the plane $\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) - b = 0$ be $Q(-3, 5, 2)$. Then the value of $|a + b|$ is equal to ......

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