If a plane passes through the points (-1, k, | vedclass

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(D) The given line is $\frac{x-1}{1} = \frac{2y+1}{2} = \frac{z+1}{-1}$.Rewriting the line equation: $\frac{x-1}{1} = \frac{y+1/2}{1} = \frac{z+1}{-1}

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$\frac{13}{6}$

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(D) The given line is $\frac{x-1}{1} = \frac{2y+1}{2} = \frac{z+1}{-1}$.Rewriting the line equation: $\frac{x-1}{1} = \frac{y+1/2}{1} = \frac{z+1}{-1}$.The direction vector of the line is $\vec{v} = \hat{i} + \hat{j} - \hat{k}$.Let the points be $A(-1, k, 0), B(2, k, -1), C(1, 1, 2)$.Vectors in the plane are $\vec{CA} = (-1-1)\hat{i} + (k-1)\hat{j} + (0-2)\hat{k} = -2\hat{i} + (k-1)\hat{j} - 2\hat{k}$ and $\vec{CB} = (2-1)\hat{i} + (k-1)\hat{j} + (-1-2)\hat{k} = \hat{i} + (k-1)\hat{j} - 3\hat{k}$.The normal vector $\vec{n}$ to the plane is $\vec{CA} 	imes \vec{CB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2 & k-1 & -2 \ 1 & k-1 & -3 \end{vmatrix}$.$\vec{n} = \hat{i}(-3(k-1) + 2(k-1)) - \hat{j}(6 + 2) + \hat{k}(-2(k-1) - (k-1)) = -(k-1)\hat{i} - 8\hat{j} - 3(k-1)\hat{k}$.Since the plane is parallel to the line,the normal vector $\vec{n}$ is perpendicular to the line's direction vector $\vec{v}$.Thus,$\vec{n} \cdot \vec{v} = 0 \Rightarrow 1(-(k-1)) + 1(-8) - 1(-3(k-1)) = 0$.$-k + 1 - 8 + 3k - 3 = 0 \Rightarrow 2k - 10 = 0 \Rightarrow k = 5$.Substituting $k=5$ into the expression: $\frac{k^2+1}{(k-1)(k-2)} = \frac{5^2+1}{(5-1)(5-2)} = \frac{26}{4 	imes 3} = \frac{26}{12} = \frac{13}{6}$.

If a plane passes through the points $(-1, k, 0), (2, k, -1), (1, 1, 2)$ and is parallel to the line $\frac{x-1}{1} = \frac{2y+1}{2} = \frac{z+1}{-1}$,then the value of $\frac{k^2+1}{(k-1)(k-2)}$ is

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