The distance of the point (1, 2, 0) from the line \(\displaystyle\frac{x-3}{2} = \frac{y+4}{3} = \frac{z+6}{5}\) measured parallel to the plane x + y + z = 3 is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\sqrt{26}\)
Let P = (1, 2, 0). A general point Q of the line is
\(Q = (3 + 2t,\ -4 + 3t,\ -6 + 5t)\)
so
\(\vec{PQ} = (2 + 2t)\hat{\imath} + (-6 + 3t)\hat{\jmath} + (-6 + 5t)\hat{k}\)
The distance is measured along PQ parallel to the plane, so PQ is perpendicular to the normal (1, 1, 1) of the plane:
\((2 + 2t) + (-6 + 3t) + (-6 + 5t) = 0\)
\(10t - 10 = 0,\quad t = 1\)
Then Q = (5, −1, −1) and \(\vec{PQ} = 4\hat{\imath} - 3\hat{\jmath} - \hat{k}\), so
\(PQ = \sqrt{16 + 9 + 1} = \sqrt{26}\)
Note: This topic was not in the CUET 2026 syllabus.