If a plane passes through the points (-1, k, 0), (2, k, -1), (1, 1, 2) and is parallel to the line x−11=2y+12=z+1−1, then the value of k2+1(k−1)(k−2) is

Equation of plane is x−1y−1z−211−130−1=0⇒x + 2y + 3z = 9(-1, k, 0) is on the plane ⇒k = 5k2+1(k−1)(k−2)=264×3=136

If a plane passes through the points (-1, k, 0), (2, k, -1), (1, 1, 2) and is parallel to the line x−11=2y+12=z+1−1, then the value of k2+1(k−1)(k−2) is

Equation of plane is x−1y−1z−211−130−1=0⇒x + 2y + 3z = 9(-1, k, 0) is on the plane ⇒k = 5k2+1(k−1)(k−2)=264×3=136

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If a plane passes through the points $(- 1, k, 0), (2, k, - 1), (1, 1, 2)$ and is parallel to the line $\frac{x - 1}{1} = \frac{2 y + 1}{2} = \frac{z + 1}{- 1}$ , then the value of $\frac{k^{2} + 1}{(k - 1) (k - 2)}$ is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

5/17

13/6

6/13

17/5

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Equation of plane is $|\begin{array}{ccc} x - 1 & y - 1 & z - 2 \\ 1 & 1 & - 1 \\ 3 & 0 & - 1 \end{array}| = 0$
$\Rightarrow$ x + 2y + 3z = 9
(-1, k, 0) is on the plane $\Rightarrow$ k = 5
$\frac{k^{2} + 1}{(k - 1) (k - 2)} = \frac{26}{4 \times 3} = \frac{13}{6}$