Let P_1 be the plane 3x - y - 7z = 11 and P_2 | vedclass

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(A) The equation of plane $P_2$ passing through $(2, -1, 0)$,$(2, 0, -1)$,and $(5, 1, 1)$ is given by the determinant equation:$\begin{vmatrix} x-5 &

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$11$

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(A) The equation of plane $P_2$ passing through $(2, -1, 0)$,$(2, 0, -1)$,and $(5, 1, 1)$ is given by the determinant equation:$\begin{vmatrix} x-5 & y-1 & z-1 \ 2-5 & 0-1 & -1-1 \ 2-5 & -1-1 & 0-1 \end{vmatrix} = 0 \implies \begin{vmatrix} x-5 & y-1 & z-1 \ -3 & -1 & -2 \ -3 & -2 & -1 \end{vmatrix} = 0$Expanding this,we get $(x-5)(1-4) - (y-1)(3-6) + (z-1)(6-3) = 0 \implies -3(x-5) + 3(y-1) + 3(z-1) = 0 \implies -x+5+y-1+z-1 = 0 \implies x-y-z = 3$.The direction ratios of the line of intersection of $P_1: 3x - y - 7z = 11$ and $P_2: x - y - z = 3$ are given by the cross product of their normals $\vec{n_1} = (3, -1, -7)$ and $\vec{n_2} = (1, -1, -1)$:$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -1 & -7 \ 1 & -1 & -1 \end{vmatrix} = \hat{i}(1-7) - \hat{j}(-3+7) + \hat{k}(-3+1) = -6\hat{i} - 4\hat{j} - 2\hat{k}$.We can take the direction ratios as $(3, 2, 1)$.To find a point on the line,set $z=0$: $3x-y=11$ and $x-y=3$. Subtracting gives $2x=8 \implies x=4$,so $y=1$. The point is $(4, 1, 0)$.The line is $\frac{x-4}{3} = \frac{y-1}{2} = \frac{z}{1} = r$.Any point on the line is $(3r+4, 2r+1, r)$.The vector from $(7, 4, -1)$ to this point is $(3r-3, 2r-3, r+1)$.Since this vector is perpendicular to the line direction $(3, 2, 1)$,their dot product is zero:$3(3r-3) + 2(2r-3) + 1(r+1) = 0 \implies 9r-9 + 4r-6 + r+1 = 0 \implies 14r - 14 = 0 \implies r=1$.The foot of the perpendicular is $(3(1)+4, 2(1)+1, 1) = (7, 3, 1)$.Thus,$\alpha+\beta+\gamma = 7+3+1 = 11$.

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