(A) If the line x - 4 1 = y - 2 1 = z - 2 k 2 lies in the plane 2 x - 4 y + z = 3 , then k is equal to (I) 9 2 (B) If the lines x - 1 2 = y + 1 3 = z - 1 4 and x - 3 1 = y - k 2 = z 1 intersect, then value of k is (II) 3 2 (C) If a plane pass

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Solution

Q.

List-I List-II

(A) If the line $\frac{x - 4}{1} = \frac{y - 2}{1} = \frac{z - 2 k}{2}$ lies in the plane
$2 x - 4 y + z = 3$ , then $k$ is equal to (I) $\frac{9}{2}$

(B) If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and
$\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect, then value of $k$ is (II) $\frac{3}{2}$

(C) If a plane passes through the point (1,1,1) cuts the coordinate axes at A, B & C respectively such that OA = OB = OC where O is the origin, then volume of tetrahedron OABC is equal to (III) $\frac{5}{2}$

(D) If a plane P passes through (1, -2, 1) is perpendicular to two planes
$2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ . The distance of plane $P$ from the point $(- 1, \frac{5}{\sqrt{2}}, \frac{3}{\sqrt{2}})$ is (IV) $\frac{7}{2}$

1 (A) - II, (B) - IV, (C) - I, (D) - III

2 (A) - I, (B) - III, (C) - IV, (D) - II

3 (A) - IV, (B) - II, (C) - I, (D) - III

4 (A) - II, (B) - I, (C) - III, (D) - IV

Ans.
(1)

(A) As line is parallel to the plane, the point $(4, 2, 2 k)$ should lie in the given plane

$∴$ $2 (4) - 4 (2) + 2 k = 3 \Rightarrow k = \frac{3}{2}$

(B) Any point on given lines are $(2 λ + 1, 3 λ - 1, 4 λ + 1)$ and $(μ + 3, 2 μ + k, μ)$ respectively.

By equating the corresponding coordinates $\Rightarrow k = \frac{9}{2}$

(C) Equation of plane is $\frac{x}{a} + \frac{y}{a} + \frac{z}{a} = 1$ satisfy it by given point $\Rightarrow \frac{1}{a} + \frac{1}{a} + \frac{1}{a} = 1$ $\Rightarrow a = 3$

Hence A(3,0,0), B(0,3,0), C(0,0,3) $\Rightarrow$ volume of tetrahedron $= \frac{1}{6}$ $| \begin{matrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{matrix} |$ $= \frac{9}{2} cubic units$

(D) The equation of plane is $a (x - 1) + b (y + 2) + c (z - 1) = 0$ and $2 a - 2 b + c = 0$

$a - b + 2 c = 0$ $\Rightarrow \frac{a}{1} = \frac{b}{1} = \frac{c}{0}$ Hence equation of plane is $x + y + 1 = 0$

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