A plane ( pi ) passing through the point (1, | vedclass

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(C) Let the equation of the plane be $\pi: ax + by + cz + 1 = 0$. Since it passes through $(1, 2, -3)$, we have $a(1) + b(2) + c(-3) + 1 = 0$, which i

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$2$

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(C) Let the equation of the plane be $\pi: ax + by + cz + 1 = 0$. Since it passes through $(1, 2, -3)$, we have $a(1) + b(2) + c(-3) + 1 = 0$, which implies $a + 2b - 3c = -1$ (Equation $i$). Since the plane is perpendicular to $x + y - z + 4 = 0$ and $2x - y + z + 1 = 0$, its normal vector $\vec{n} = (a, b, c)$ is perpendicular to the normal vectors of these planes, $\vec{n_1} = (1, 1, -1)$ and $\vec{n_2} = (2, -1, 1)$. Thus, $a + b - c = 0$ (Equation $ii$) and $2a - b + c = 0$ (Equation $iii$). Adding equations $(ii)$ and $(iii)$, we get $3a = 0$, so $a = 0$. Substituting $a = 0$ into $(ii)$, we get $b - c = 0$, so $b = c$. Substituting $a = 0$ and $b = c$ into $(i)$, we get $0 + 2c - 3c = -1$, which implies $-c = -1$, so $c = 1$. Thus, $b = 1$. The values are $a = 0, b = 1, c = 1$. Therefore, $a^2 + b^2 + c^2 = 0^2 + 1^2 + 1^2 = 2$.

$A$ plane ( $\pi$ ) passing through the point $(1, 2, -3)$ is perpendicular to the planes $x + y - z + 4 = 0$ and $2x - y + z + 1 = 0$. If the equation of the plane ( $\pi$ ) is $ax + by + cz + 1 = 0$, then $a^2 + b^2 + c^2 =$

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