The equation of plane that contains line \(\displaystyle\frac{x-1}{-1}=\frac{y+1}{2}=\frac{z-1}{3}\) and also passes through point (0, 1, 0) is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(2x+y-1 = 0\)
The line passes through A(1, −1, 1) with direction (−1, 2, 3). The plane also contains P(0, 1, 0), so it contains the vector
\(\overrightarrow{PA} = (1, -2, 1)\)
The normal to the plane is perpendicular to both, so it is their cross product:
\( \displaystyle \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -1 & 2 & 3 \\ 1 & -2 & 1 \end{vmatrix} = (2+6)\hat{\imath} - (-1-3)\hat{\jmath} + (2-2)\hat{k} \)
\( = 8\hat{\imath} + 4\hat{\jmath} \)
So the normal can be taken as (2, 1, 0). The plane through P(0, 1, 0):
\(2(x-0) + 1(y-1) + 0(z-0) = 0\)
\(2x + y - 1 = 0\)
Check with A(1, −1, 1): 2 − 1 − 1 = 0.
Note: This topic was not in the CUET 2026 syllabus.