The image of the point (1, 3, 4) in the plane 2x − y + z + 3 = 0 is

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(−3, 5, 2) \text{ Let Q be the image of the point P(1, 3, 4) in the plane 2x - y + z + 3 = 0 } \text{ Then PQ is normal to the plane. So, the direction ratios of PQ are proportional to 2, -1, 1.} \text{ Since PQ passes through P (1, 3, 4) and has the direction ratios proportional to 2, -1, 1 . , equation of PQ is} x - $\frac{1}{2}$ = y - 3/- 1 = z - $\frac{4}{1}$ = r (\text{ say } ) \text{ Let the coordinates of Q be } \left( 2r + 1, - r + 3, r + 4 \right). \text{ Let R be the mid point of PQ. Then } , R = \left( 2r + 1 + $\frac{1}{2}$, - r + 3 + $\frac{3}{2}$, r + 4 + $\frac{4}{2}$ \right) = \left( r + 1, - r + $\frac{6}{2}$, r + $\frac{8}{2}$ \right) \text{ Since R lies in the plane } 2x - y + z + 3 = 0, 2 \left( r + 1 \right) - \left( - r + $\frac{6}{2}$ \right) + r + $\frac{8}{2}$ + 3 = 0 \Rightarrow 4r + 4 + r - 6 + r + 8 + 6 = 0 \Rightarrow 6r + 12 = 0 \Rightarrow r = - 2 \text{ Substituting this in the coordinates of Q, we get } Q = \left( 2r + 1, - r + 3, r + 4 \right) . = \left( 2 \left( - 2 \right) + 1, 2 + 3, - 2 + 4 \right) = \left( - 3, 5, 2 \right) So, the answer is (b).

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