The point of intersection of the lines l_1: r | vedclass

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(C) Let the point of intersection be $P$. Any point on line $l_1$ is given by $(1+t, -6+2t, 2+t)$.Any point on line $l_2$ is given by $(2u, 4+u, 1+2u)

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$(8, 8, 9)$

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(C) Let the point of intersection be $P$. Any point on line $l_1$ is given by $(1+t, -6+2t, 2+t)$.Any point on line $l_2$ is given by $(2u, 4+u, 1+2u)$.For the intersection point,we equate the coordinates:$1+t = 2u$  $(i)$$-6+2t = 4+u$  $(ii)$$2+t = 1+2u$  $(iii)$From $(i)$,$t = 2u - 1$. Substituting this into $(ii)$:$-6 + 2(2u - 1) = 4 + u$$-6 + 4u - 2 = 4 + u$$3u = 12 \Rightarrow u = 4$.Substituting $u = 4$ into $t = 2u - 1$,we get $t = 2(4) - 1 = 7$.Now,find the point $P$ using $t = 7$ in $l_1$:$P = (1+7, -6+2(7), 2+7) = (8, 8, 9)$.

The point of intersection of the lines $l_1: r(t) = (i - 6j + 2k) + t(i + 2j + k)$ and $l_2: R(u) = (4j + k) + u(2i + j + 2k)$ is

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