Key Concepts and Formulas

1. Family of Planes: If $P_1: A_1x + B_1y + C_1z + D_1 = 0$ and $P_2: A_2x + B_2y + C_2z + D_2 = 0$ are two distinct planes, then the equation of any plane passing through their line of intersection is given by $P_1 + \lambda P_2 = 0$, where $\lambda$ is a scalar constant. This equation can be expanded to $(A_1 + \lambda A_2)x + (B_1 + \lambda B_2)y + (C_1 + \lambda C_2)z + (D_1 + \lambda D_2) = 0$.

2. Condition for a Plane to be Parallel to an Axis: A plane $Ax + By + Cz + D = 0$ is parallel to a coordinate axis if its normal vector $\vec{n} = \langle A, B, C \rangle$ is perpendicular to the direction vector of that axis.

   • If the plane is parallel to the x-axis, the coefficient of $x$ (i.e., $A$) must be zero.
   • If the plane is parallel to the y-axis, the coefficient of $y$ (i.e., $B$) must be zero.
   • If the plane is parallel to the z-axis, the coefficient of $z$ (i.e., $C$) must be zero.

Step-by-Step Solution

Step 1: Formulate the general equation of the plane passing through the intersection We are given the equations of two planes: Plane 1 ($P_1$): $x + y + z = 1 \implies x + y + z - 1 = 0$ Plane 2 ($P_2$): $2x + 3y - z + 4 = 0$

According to the concept of a family of planes, the equation of any plane passing through the line of intersection of $P_1$ and $P_2$ is given by $P_1 + \lambda P_2 = 0$. Substituting the equations of $P_1$ and $P_2$: $(x + y + z - 1) + \lambda (2x + 3y - z + 4) = 0$ To make it easier to work with, we group the terms by $x$, $y$, and $z$: $(1 + 2\lambda)x + (1 + 3\lambda)y + (1 - \lambda)z + (4\lambda - 1) = 0$ This is the general equation of a plane that satisfies the first condition (passing through the line of intersection).

Step 2: Apply the condition for parallelism to the y-axis The problem states that the required plane is parallel to the y-axis. As per the key concepts, for a plane $Ax + By + Cz + D = 0$ to be parallel to the y-axis, its normal vector $\vec{n} = \langle A, B, C \rangle$ must be perpendicular to the direction vector of the y-axis, $\hat{j} = \langle 0, 1, 0 \rangle$. This implies that the coefficient of $y$ in the plane's equation must be zero.

From our general plane equation derived in Step 1, the coefficient of $y$ is $(1 + 3\lambda)$. Setting this to zero: $1 + 3\lambda = 0$

Step 3: Calculate the value of $\lambda$ From the condition established in Step 2: $1 + 3\lambda = 0$ $3\lambda = -1$ $\lambda = -\frac{1}{3}$ This unique value of $\lambda$ defines the specific plane that fulfills both conditions: passing through the intersection of the two given planes and being parallel to the y-axis.

Step 4: Determine the equation of the specific plane Now, substitute the value $\lambda = -\frac{1}{3}$ back into the general equation of the plane from Step 1: $\left(1 + 2\left(-\frac{1}{3}\right)\right)x + \left(1 + 3\left(-\frac{1}{3}\right)\right)y + \left(1 - \left(-\frac{1}{3}\right)\right)z + \left(4\left(-\frac{1}{3}\right) - 1\right) = 0$ Let's simplify each coefficient:

• Coefficient of $x$: $1 - \frac{2}{3} = \frac{1}{3}$
• Coefficient of $y$: $1 - 1 = 0$ (This confirms our condition for parallelism to the y-axis)
• Coefficient of $z$: $1 + \frac{1}{3} = \frac{4}{3}$
• Constant term: $-\frac{4}{3} - 1 = -\frac{4}{3} - \frac{3}{3} = -\frac{7}{3}$

So, the equation of the plane is: $\frac{1}{3}x + 0y + \frac{4}{3}z - \frac{7}{3} = 0$ To eliminate the fractions and simplify, multiply the entire equation by 3: $x + 4z - 7 = 0$ This is the equation of the required plane.

Step 5: Verify which point lies on the plane We need to check which of the given options satisfies the equation of the plane $x + 4z - 7 = 0$. We will substitute the coordinates $(x, y, z)$ of each option into the equation.

• (A) (–3, 0, -1) Substitute $x = -3$ and $z = -1$: $(-3) + 4(-1) - 7 = -3 - 4 - 7 = -14$ Since $-14 \neq 0$, point (A) does not lie on the plane.

• (B) (3, 2, 1) Substitute $x = 3$ and $z = 1$: $(3) + 4(1) - 7 = 3 + 4 - 7 = 7 - 7 = 0$ Since $0 = 0$, point (B) lies on the plane.

• (C) (3, 3, -1) Substitute $x = 3$ and $z = -1$: $(3) + 4(-1) - 7 = 3 - 4 - 7 = -1 - 7 = -8$ Since $-8 \neq 0$, point (C) does not lie on the plane.

• (D) (–3, 1, 1) Substitute $x = -3$ and $z = 1$: $(-3) + 4(1) - 7 = -3 + 4 - 7 = 1 - 7 = -6$ Since $-6 \neq 0$, point (D) does not lie on the plane.

From the verification, only point (B) $(3, 2, 1)$ satisfies the equation of the derived plane.

Common Mistakes & Tips

• Algebraic Precision: Be extremely careful with signs and fractions during substitution and simplification. A minor error can propagate and lead to an incorrect plane equation.
• Understanding Parallelism: Clearly recall that a plane parallel to a coordinate axis means the coefficient of that axis's variable in the plane equation is zero.
• Systematic Verification: Always test all given options carefully. Do not assume the first option you test is correct or incorrect without performing the full calculation.

Summary This problem involved finding the equation of a plane that satisfies two conditions: passing through the line of intersection of two given planes and being parallel to the y-axis. We used the family of planes concept to set up a general equation involving a parameter $\lambda$. The condition for parallelism to the y-axis allowed us to determine the specific value of $\lambda$. Substituting this value back into the general equation yielded the unique equation of the required plane, which was $x + 4z - 7 = 0$. Finally, we checked which of the given points lies on this plane by substituting their coordinates. The point $(3, 2, 1)$ was found to satisfy the plane equation.

Final Answer The final answer is $\boxed{(3, 2, 1)}$, which corresponds to option (B).