Question

The equation of the plane passing through the point (1, 2, -3) and perpendicular to the planes $3x + y - 2z = 5$ and $2x - 5y - z = 7$, is :

• A. $6x - 5y + 2z + 10 = 0$
• B. $11x + y + 17z + 38 = 0$
• C. $6x - 5y - 2z - 2 = 0$
• D. $3x - 10y - 2z + 11 = 0$

Hint:

To find a vector perpendicular to two given vectors, always use the cross product.

Answer 1

The Correct Option is B

Step 1: The normal to the required plane is \( \vec{n} = \vec{n}_1 \times \vec{n}_2 \).

Step 2: \[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 2 & -5 & -1 \end{vmatrix} = \hat{i}(-1 - 10) - \hat{j}(-3 + 4) + \hat{k}(-15 - 2) = -11\hat{i} - \hat{j} - 17\hat{k}. \]
Step 3: Plane equation: \[ -11(x - 1) - (y - 2) - 17(z + 3) = 0. \]
Step 4: Simplifying, we get: \[ 11x + y + 17z + 38 = 0. \]