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(B) Step $1$: Solve the system of linear equations $2a + b - c = 4$, $-a + 2b + c = 0$, and $a - b + 2c = -2$.Step $2$: Adding the three equations: $(

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$3 	ext{ units.}$

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(B) Step $1$: Solve the system of linear equations $2a + b - c = 4$, $-a + 2b + c = 0$, and $a - b + 2c = -2$.Step $2$: Adding the three equations: $(2a-a+a) + (b+2b-b) + (-c+c+2c) = 4+0-2 \implies 2a + 2b + 2c = 2 \implies a + b + c = 1$.Step $3$: From $-a + 2b + c = 0$, we have $c = a - 2b$. Substitute into $a+b+c=1$: $a+b+(a-2b)=1 \implies 2a-b=1 \implies b=2a-1$.Step $4$: Substitute $b$ and $c$ into $2a+b-c=4$: $2a + (2a-1) - (a - 2(2a-1)) = 4 \implies 2a + 2a - 1 - a + 4a - 2 = 4 \implies 7a = 7 \implies a=1$.Step $5$: Then $b = 2(1)-1 = 1$ and $c = 1-2(1) = -1$. Point $P$ is $(1, 1, -1)$.Step $6$: Distance from $P(1, 1, -1)$ to $2x + y + 2z - 10 = 0$ is $d = \frac{|2(1) + 1(1) + 2(-1) - 10|}{\sqrt{2^2 + 1^2 + 2^2}} = \frac{|2 + 1 - 2 - 10|}{\sqrt{9}} = \frac{|-9|}{3} = 3 	ext{ units.}$

If $\begin{bmatrix} 2 & 1 & -1 \\ -1 & 2 & 1 \\ 1 & -1 & 2 \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} = \begin{bmatrix} 4 \\ 0 \\ -2 \end{bmatrix}$, then the distance of point $P(a, b, c)$ from the plane $2x + y + 2z = 10$ is

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