Let the line L pass through the point (0,1,2) | vedclass

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(D) Let the line $L$ pass through $A(0,1,2)$ and intersect the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ at point $B(1+2\lambda, 2+3\lambda, 3+

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$\sqrt{74}$

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(D) Let the line $L$ pass through $A(0,1,2)$ and intersect the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ at point $B(1+2\lambda, 2+3\lambda, 3+4\lambda)$.The direction vector of line $L$ is $\vec{v} = \vec{AB} = (1+2\lambda)\hat{i} + (1+3\lambda)\hat{j} + (1+4\lambda)\hat{k}$.Since $L$ is parallel to the plane $2x+y-3z=4$,the normal vector $\vec{n} = 2\hat{i} + \hat{j} - 3\hat{k}$ is perpendicular to $\vec{v}$.Thus,$\vec{v} \cdot \vec{n} = 0 \Rightarrow 2(1+2\lambda) + 1(1+3\lambda) - 3(1+4\lambda) = 0$.$2 + 4\lambda + 1 + 3\lambda - 3 - 12\lambda = 0 \Rightarrow -5\lambda = 0 \Rightarrow \lambda = 0$.So,the point $B$ is $(1, 2, 3)$ and the direction vector $\vec{v} = \hat{i} + \hat{j} + \hat{k}$.The equation of line $L$ is $\vec{r} = (0\hat{i} + 1\hat{j} + 2\hat{k}) + t(\hat{i} + \hat{j} + \hat{k})$.Let $Q$ be the projection of $P(1,-9,2)$ on $L$. $Q = (t, 1+t, 2+t)$.$\vec{PQ} = (t-1)\hat{i} + (10+t)\hat{j} + t\hat{k}$.Since $\vec{PQ} \perp \vec{v}$,$\vec{PQ} \cdot \vec{v} = 0 \Rightarrow (t-1) + (10+t) + t = 0 \Rightarrow 3t = -9 \Rightarrow t = -3$.$Q = (-3, -2, -1)$.Distance $PQ = \sqrt{(-3-1)^2 + (-2 - (-9))^2 + (-1-2)^2} = \sqrt{(-4)^2 + 7^2 + (-3)^2} = \sqrt{16 + 49 + 9} = \sqrt{74}$.

Let the line $L$ pass through the point $(0,1,2)$,intersect the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and be parallel to the plane $2x+y-3z=4$. Then the distance of the point $P(1,-9,2)$ from the line $L$ is

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