Given equation of the plane is \(5x+2y-3z=17\). Then,
A. point (2, 2, −1) lies on the plane
B. The direction cosines of normal to plane are (2, 2, −1)
C. Coordinates of foot of perpendicular from origin are (5, 2, −3)
D. The plane does not pass through the origin
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A and D only
The normal to \(5x + 2y - 3z = 17\) has direction ratios 5, 2, −3, and
\(\sqrt{5^2 + 2^2 + (-3)^2} = \sqrt{38}\)
so its direction cosines are \(\frac{5}{\sqrt{38}}, \frac{2}{\sqrt{38}}, \frac{-3}{\sqrt{38}}\). The foot of the perpendicular from the origin is the point \(t(5, 2, -3)\) on the plane:
\(25t + 4t + 9t = 17\)
\( \displaystyle t = \frac{17}{38} \)
So the direction cosines of the normal are \(\frac{5}{\sqrt{38}}, \frac{2}{\sqrt{38}}, \frac{-3}{\sqrt{38}}\) and the foot of the perpendicular is \(\left(\frac{85}{38}, \frac{34}{38}, -\frac{51}{38}\right)\).
Checking the statements
(A) point (2, 2, −1) lies on the plane — Correct. 5(2) + 2(2) − 3(−1) = 10 + 4 + 3 = 17.
(B) The direction cosines of normal to plane are (2, 2, −1) — Incorrect. They are proportional to 5, 2, −3, not to 2, 2, −1; and \(2^2 + 2^2 + 1^2 = 9 \ne 1\).
(C) Coordinates of foot of perpendicular from origin are (5, 2, −3) — Incorrect. (5, 2, −3) is not even on the plane: 25 + 4 + 9 = 38 ≠ 17.
(D) The plane does not pass through the origin — Correct. At (0, 0, 0) the left side is 0 ≠ 17.
Note: This topic was not in the CUET 2026 syllabus.