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# Find the image of the point (1, 3. 4) in the plane $x-y+z =5$. Hence show that the image lies on the plane $x -2y +z -7 = 0$

$(A)\; (3, -1, 6)$
$(B)\; (-3, 1, 6)$
$(C)\; (3,1,6)$
$(D)\; (-3, 1, -6)$

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A)
The equation of the plane is $x - y +z = 5----(1)$
Direction ratios of the normal to plane are (1, -1, 1)
Let M be the foot of the perpendicular from P(1, 3, 4) to the plane (1)
(ie) PM is perpendicular to the plane.
Hence its directed rations are (1, -1, 1) and it passes through P (1, 3, 4)
$\therefore$ Its equation is $\frac{x-1}{1} = \frac{y-3}{-1} = \frac{z-4}{1} = k$
Any point on the line is $M (K+1 ,-K+3 ,K+4)$
Since M lies on the plane :
\; \; \;\; \; \;\begin{align*} (K+1) - (-K+3) + (K+4 ) &= 5 \\ \implies 3K & = 3 \\ \implies K &= 1\end{align*}
$\therefore$Foot of the perpendicular is M (2, 2, 5)
M is ttthe mid point of P
$\therefore \frac{1 + \lambda}{2} =2 \implies \lambda = 3$
$\frac{3 + \beta }{2} = 2 \implies \beta = 1$
$\frac{4+x}{2} = 5 \implies x = 6$
Hence $P' =(3, 1, 6)$
It lies on the plane $x-2y+z -7 = 0$
\begin{align*}( ie) 3 -2(1) + 6 -7 &= 0 \\ 9 - 9 &= 0 \end{align*}
Hence $P' (3, 1, 6)$ lies on the plane $x -2y +z-7 =0$