Let frac{x-2}{3}=frac{y+1}{-2}=frac{z+3}{-1} | vedclass

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(B) The line passes through the point $(2, -1, -3)$. Since the line lies on the plane $px - qy + z = 5$,this point must satisfy the plane equation:$p(

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$\sqrt{\frac{5}{142}}$

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(B) The line passes through the point $(2, -1, -3)$. Since the line lies on the plane $px - qy + z = 5$,this point must satisfy the plane equation:$p(2) - q(-1) + (-3) = 5 \Rightarrow 2p + q = 8$  --- $(i)$The direction vector of the line is $\vec{v} = (3, -2, -1)$ and the normal to the plane is $\vec{n} = (p, -q, 1)$. Since the line lies on the plane,$\vec{v} \cdot \vec{n} = 0$:$3(p) + (-2)(-q) + (-1)(1) = 0 \Rightarrow 3p + 2q = 1$  --- $(ii)$Solving equations $(i)$ and $(ii)$:Multiply $(i)$ by $2$: $4p + 2q = 16$Subtract $(ii)$ from this: $(4p + 2q) - (3p + 2q) = 16 - 1 \Rightarrow p = 15$Substitute $p = 15$ into $(i)$: $2(15) + q = 8 \Rightarrow 30 + q = 8 \Rightarrow q = -22$The equation of the plane is $15x + 22y + z = 5$,or $15x + 22y + z - 5 = 0$.The distance $d$ from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.$d = \frac{|-5|}{\sqrt{15^2 + 22^2 + 1^2}} = \frac{5}{\sqrt{225 + 484 + 1}} = \frac{5}{\sqrt{710}} = \sqrt{\frac{25}{710}} = \sqrt{\frac{5}{142}}$.

Let $\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z+3}{-1}$ lie on the plane $px-qy+z=5$,for some $p, q \in R$. The shortest distance of the plane from the origin is

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