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A point on the plane passes through the points $(1, - 1, 6) (0, 0, 7)$ and perpendicular to the plane $x - 2 y + z = 6$ is

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$(1, - 1, 2)$

$(1, 1, 2)$

$(- 1, 1, 2)$

$(1, 1, - 2)$

answer is B.

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Detailed Solution

$\begin{array}{rcl} |\begin{matrix} x - 1 & y + 1 & z - 6 \\ - 1 & 1 & 1 \\ 1 & - 2 & 1 \end{matrix}| & = & 0 \\ (x - 1) (3) - (y + 1) (- 2) + (z - 6) (1) & = & 0 \\ 3 x - 3 + 2 y + 2 + z - 6 & = & 0 \\ 3 x + 2 y + 3 - 7 & = & 0 \end{array}$

(1, 1, 2) lies on above plane

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