A plane pi is passing through (2,0,1) and (3, | vedclass

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(C) Let the equation of the plane $\pi$ be $a(x-2) + b(y-0) + c(z-1) = 0$,which simplifies to $ax + by + cz = 2a + c$. Since the plane passes through

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$x-y-z+1=0$

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(C) Let the equation of the plane $\pi$ be $a(x-2) + b(y-0) + c(z-1) = 0$,which simplifies to $ax + by + cz = 2a + c$. Since the plane passes through $(3, -3, 4)$,we have $3a - 3b + 4c = 2a + c$,which gives $a - 3b + 3c = 0$. Since the plane $\pi$ is perpendicular to $x - 2y + z = 6$,the normal vectors are perpendicular,so $a(1) + b(-2) + c(1) = 0$,which gives $a - 2b + c = 0$. Solving the system of equations: $a - 3b + 3c = 0$ $a - 2b + c = 0$ Subtracting the equations: $(-3b - (-2b)) + (3c - c) = 0 \implies -b + 2c = 0 \implies b = 2c$. Substituting $b = 2c$ into $a - 2b + c = 0$: $a - 2(2c) + c = 0 \implies a - 3c = 0 \implies a = 3c$. Setting $c = 1$,we get $a = 3$ and $b = 2$. The equation of plane $\pi$ is $3x + 2y + z = 2(3) + 1 = 7$. $A$ plane is perpendicular to $\pi$ if its normal vector is perpendicular to the normal vector of $\pi$ $(3, 2, 1)$. Checking option $C$: The normal vector is $(1, -1, -1)$. Dot product: $(3)(1) + (2)(-1) + (1)(-1) = 3 - 2 - 1 = 0$. Since the dot product is $0$,the plane $x - y - z + 1 = 0$ is perpendicular to $\pi$.

$A$ plane $\pi$ is passing through $(2,0,1)$ and $(3,-3,4)$ and is perpendicular to the plane $x-2y+z=6$. Among the following,the plane that is perpendicular to the plane $\pi$ is:

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