Q71 / 150 · MHT CET 2026 (15 April Shift 2) · Mathematics
The equation of the plane passing through the point $(1,2,1)$ and perpendicular to the planes $x + 2y + 2z - 7 = 0$ and $3x + 3y + 2z - 5 = 0$ is
- A$2x + y - 2z - 2 = 0$
- Correct answerB$2x - 4y + 3z + 3 = 0$
- C$2x + 4y - 5z - 5 = 0$
- D$x + 4y - 6z - 3 = 0$
Solution
Let the normal vector to the required plane be $\vec{n}$.
Since the required plane is perpendicular to the given planes, its normal vector is perpendicular to the normal vectors of the given planes, $\vec{n}_1 = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{n}_2 = 3\hat{i} + 3\hat{j} + 2\hat{k}$.
$\vec{n} = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 3 & 3 & 2 \end{vmatrix}$
$\vec{n} = \hat{i}(4 - 6) - \hat{j}(2 - 6) + \hat{k}(3 - 6) = -2\hat{i} + 4\hat{j} - 3\hat{k}$
The equation of the plane passing through $(1, 2, 1)$ with normal vector $\vec{n}$ is:
$-2(x - 1) + 4(y - 2) - 3(z - 1) = 0$
$-2x + 2 + 4y - 8 - 3z + 3 = 0$
$-2x + 4y - 3z - 3 = 0$
$2x - 4y + 3z + 3 = 0$
Answer: $2x - 4y + 3z + 3 = 0$
Since the required plane is perpendicular to the given planes, its normal vector is perpendicular to the normal vectors of the given planes, $\vec{n}_1 = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{n}_2 = 3\hat{i} + 3\hat{j} + 2\hat{k}$.
$\vec{n} = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 3 & 3 & 2 \end{vmatrix}$
$\vec{n} = \hat{i}(4 - 6) - \hat{j}(2 - 6) + \hat{k}(3 - 6) = -2\hat{i} + 4\hat{j} - 3\hat{k}$
The equation of the plane passing through $(1, 2, 1)$ with normal vector $\vec{n}$ is:
$-2(x - 1) + 4(y - 2) - 3(z - 1) = 0$
$-2x + 2 + 4y - 8 - 3z + 3 = 0$
$-2x + 4y - 3z - 3 = 0$
$2x - 4y + 3z + 3 = 0$
Answer: $2x - 4y + 3z + 3 = 0$
Hence, the correct answer is option: 'B'.