(1, -2, 1) is a point on a plane pi and pi is | vedclass

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(D) Since the plane $\pi$ is parallel to the plane $x-y-z=0$, its equation must be of the form $x-y-z+k=0$. Given that the point $(1, -2, 1)$ lies on

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$a$

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(D) Since the plane $\pi$ is parallel to the plane $x-y-z=0$, its equation must be of the form $x-y-z+k=0$. Given that the point $(1, -2, 1)$ lies on the plane $\pi$, we substitute these coordinates into the equation: $1 - (-2) - 1 + k = 0$ $1 + 2 - 1 + k = 0$ $2 + k = 0 \implies k = -2$. Thus, the equation of the plane $\pi$ is $x-y-z-2=0$. Comparing this with the given equation $ax+by+cz-2=0$, we get $a=1, b=-1, c=-1$. Now, calculating $b-2c$: $b-2c = -1 - 2(-1) = -1 + 2 = 1$. Since $a=1$, we have $b-2c = a$.

$(1, -2, 1)$ is a point on a plane $\pi$ and $\pi$ is parallel to the plane $x-y-z=0$. If the equation of $\pi$ is $ax+by+cz-2=0$, then $b-2c=$

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