I'm drawing a blank as to how to proceed.
2026-04-29 07:35:26.1777448126
Find the line `(t) through (3, 1, −2) that intersects and is perpendicular to the line s(t) = (−1 + t, −2 + t, −1 + t).
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Let $A(3,1,-2)$ and $B(-1+t,-2+t,-1+t)$.
Thus, $$\vec{AB}(-4+t,-3+t,1+t)$$ and since we need $AB\perp(1,1,1),$ we obtain $$-4+t-3+t+1+t=0,$$ which gives $t=2$, $$\vec{AB}(-2,-1,3)$$ and we got the answer: $$(3-2s,1-s,-2+3s).$$