Let the plane passing through the point (2,1, | vedclass

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(A) The line passes through $A(1,3,2)$ and $B(1,2,1)$. The direction vector of the line is $\vec{v} = B - A = (1-1, 2-3, 1-2) = (0, -1, -1)$.The plane

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(A) The line passes through $A(1,3,2)$ and $B(1,2,1)$. The direction vector of the line is $\vec{v} = B - A = (1-1, 2-3, 1-2) = (0, -1, -1)$.The plane passes through $P(2,1,-1)$ and $A(1,3,2)$. The vector $\vec{PA} = (1-2, 3-1, 2-(-1)) = (-1, 2, 3)$.The normal to the plane is $\vec{n} = \vec{v} 	imes \vec{PA} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & -1 & -1 \ -1 & 2 & 3 \end{vmatrix} = \hat{i}(-3+2) - \hat{j}(0-1) + \hat{k}(0-1) = -\hat{i} + \hat{j} - \hat{k}$.The equation of the plane is $-1(x-2) + 1(y-1) - 1(z+1) = 0$,which simplifies to $-x+2+y-1-z-1 = 0$,or $-x+y-z = 0$,i.e.,$x-y+z = 0$.Since the plane passes through the origin $(0,0,0)$,the intercepts $p, q, r$ are all $0$.Therefore,$p+q+r = 0+0+0 = 0$.

Let the plane passing through the point $(2,1,-1)$ and containing the line joining the points $(1,3,2)$ and $(1,2,1)$ make intercepts $p, q, r$ on the coordinate axes. Then $p+q+r=$

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