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(C) Let the general point $P$ on the line $\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-1}{-1}=r$ be $P(2r+1, r-1, 1-r)$.Since $P$ lies on the plane $x+2y+3z=4

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$\frac{x-5}{-3}=\frac{y-1}{-2}=\frac{z+1}{-7}$

Vedclass · Answer

(C) Let the general point $P$ on the line $\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-1}{-1}=r$ be $P(2r+1, r-1, 1-r)$.Since $P$ lies on the plane $x+2y+3z=4$,we substitute the coordinates of $P$ into the plane equation:$(2r+1) + 2(r-1) + 3(1-r) = 4$$2r + 1 + 2r - 2 + 3 - 3r = 4$$r + 2 = 4 \Rightarrow r = 2$.Substituting $r=2$ into the coordinates of $P$,we get the point of intersection as $P(5, 1, -1)$.The direction vector of the required line is given by the cross product:$\vec{v} = (2\hat{i}-3\hat{j}) 	imes (\hat{i}+2\hat{j}-\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -3 & 0 \ 1 & 2 & -1 \end{vmatrix}$$= \hat{i}(3-0) - \hat{j}(-2-0) + \hat{k}(4+3) = 3\hat{i} + 2\hat{j} + 7\hat{k}$.The equation of the line passing through $(5, 1, -1)$ with direction vector $(3, 2, 7)$ is $\frac{x-5}{3} = \frac{y-1}{2} = \frac{z+1}{7}$.Multiplying the denominators by $-1$,we get $\frac{x-5}{-3} = \frac{y-1}{-2} = \frac{z+1}{-7}$,which matches option $C$.

The equation of the line passing through the intersection of the plane $x+2y+3z=4$ and the line $\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-1}{-1}$ and parallel to the vector $(2\hat{i}-3\hat{j}) \times (\hat{i}+2\hat{j}-\hat{k})$ is

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