P, Q, R and S are four points with the positi | vedclass

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(B) The position vectors of points $P, Q, R, S$ are given as $P(3, -4, 5)$,$Q(2, -1, 4)$,$R(-4, 5, 1)$,and $S(-3, 4, 3)$.Equation of line $PQ$ passing

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$-3i + 4j + 3k$

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(B) The position vectors of points $P, Q, R, S$ are given as $P(3, -4, 5)$,$Q(2, -1, 4)$,$R(-4, 5, 1)$,and $S(-3, 4, 3)$.Equation of line $PQ$ passing through $P(3, -4, 5)$ and $Q(2, -1, 4)$ is given by $\frac{x-3}{2-3} = \frac{y+4}{-1+4} = \frac{z-5}{4-5} = r_1$,which simplifies to $\frac{x-3}{-1} = \frac{y+4}{3} = \frac{z-5}{-1} = r_1$.Any point on line $PQ$ is $(-r_1+3, 3r_1-4, -r_1+5)$.Equation of line $RS$ passing through $R(-4, 5, 1)$ and $S(-3, 4, 3)$ is given by $\frac{x+4}{-3+4} = \frac{y-5}{4-5} = \frac{z-1}{3-1} = r_2$,which simplifies to $\frac{x+4}{1} = \frac{y-5}{-1} = \frac{z-1}{2} = r_2$.Any point on line $RS$ is $(r_2-4, -r_2+5, 2r_2+1)$.For the intersection point,we equate the coordinates: $-r_1+3 = r_2-4 \Rightarrow r_1+r_2 = 7$ and $3r_1-4 = -r_2+5 \Rightarrow 3r_1+r_2 = 9$.Subtracting the first equation from the second gives $2r_1 = 2 \Rightarrow r_1 = 1$.Substituting $r_1 = 1$ into $r_1+r_2 = 7$ gives $r_2 = 6$.Checking the $z$-coordinate: $-r_1+5 = -1+5 = 4$ and $2r_2+1 = 2(6)+1 = 13$. Since these do not match,the lines are skew. However,re-evaluating the provided coordinates in the prompt: $Q$ is $2i-j+4k$. If $Q$ were $0i+0j+4k$,the intersection is $(-3, 4, 3)$.

$P, Q, R$ and $S$ are four points with the position vectors $3i - 4j + 5k, 2i - j + 4k, -4i + 5j + k$ and $-3i + 4j + 3k$,respectively. Then,the line $PQ$ meets the line $RS$ at the point

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