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The equation of plane that contains line x−1−1=y+12=z−13\frac{x - 1}{-1} = \frac{y + 1}{2} = \frac{z - 1}{3} and also pass through point (0,1,0)(0, 1, 0) is :

Solution

✅ Correct Option: 1

Line passes through (1,−1,1)(1, -1, 1) with direction d⃗=(−1,2,3)\vec{d} = (-1, 2, 3).

Vector from (0,1,0)(0,1,0) to (1,−1,1)(1,-1,1): v⃗=(1,−2,1)\vec{v} = (1, -2, 1).

Normal: n⃗=v⃗×d⃗=(−6−2,−3−1,2−2)=(−8,−4,0)\vec{n} = \vec{v} \times \vec{d} = (-6-2, -3-1, 2-2) = (-8, -4, 0).

Simplify by −4-4: (2,1,0)(2, 1, 0).

Plane: 2(x−0)+1(y−1)+0(z)=0⇒2x+y−1=02(x-0) + 1(y-1) + 0(z) = 0 \Rightarrow 2x + y - 1 = 0.

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