Q60 / 150 · MHT CET 2026 (16 April Shift 1) · Mathematics
If the product of the distances of the point (1, 2, 3) from the origin and the plane $2x - 3y + z + k = 0$ is 7, then the value of k is
- Correct answerA$8$
- B$10$
- C$7$
- D$5$
Solution
Distance of the point $(1, 2, 3)$ from the origin is $d_1 = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}$
Distance of the point $(1, 2, 3)$ from the plane $2x - 3y + z + k = 0$ is $d_2 = \dfrac{|2(1) - 3(2) + 1(3) + k|}{\sqrt{2^2 + (-3)^2 + 1^2}} = \dfrac{|k - 1|}{\sqrt{14}}$
Given that $d_1 \times d_2 = 7$
$\sqrt{14} \times \dfrac{|k - 1|}{\sqrt{14}} = 7$
$|k - 1| = 7$
$k - 1 = 7$ or $k - 1 = -7$
$k = 8$ or $k = -6$
From the given options, $k = 8$.
Answer: $8$
Distance of the point $(1, 2, 3)$ from the plane $2x - 3y + z + k = 0$ is $d_2 = \dfrac{|2(1) - 3(2) + 1(3) + k|}{\sqrt{2^2 + (-3)^2 + 1^2}} = \dfrac{|k - 1|}{\sqrt{14}}$
Given that $d_1 \times d_2 = 7$
$\sqrt{14} \times \dfrac{|k - 1|}{\sqrt{14}} = 7$
$|k - 1| = 7$
$k - 1 = 7$ or $k - 1 = -7$
$k = 8$ or $k = -6$
From the given options, $k = 8$.
Answer: $8$
Hence, the correct answer is option: 'A'.