1. Key Concepts and Formulas

• Foot of the Perpendicular from a Point to a Plane: Given a point $P(x_1, y_1, z_1)$ and a plane $Ax+By+Cz+D=0$, the foot of the perpendicular $F(x_f, y_f, z_f)$ from $P$ to the plane lies on the line passing through $P$ and having the direction ratios of the plane's normal vector $(A,B,C)$. The coordinates of $F$ can be found using the formula: $\frac{x_f - x_1}{A} = \frac{y_f - y_1}{B} = \frac{z_f - z_1}{C} = \lambda$ where $\lambda$ is a parameter.
• Point on a Circle: A point $(x_0, y_0, z_0)$ lies on the circle $x^2+y^2=r^2, z=k$ if its coordinates satisfy both $x_0^2+y_0^2=r^2$ and $z_0=k$.
• Point on a Plane: A point $(x,y,z)$ lies on the plane $Ax+By+Cz+D=0$ if its coordinates satisfy the plane's equation.

2. Step-by-Step Solution

Step 1: Identify the given information and define variables. Let $P(x_0, y_0, 0)$ be a point on the given circle $x^2+y^2=1, z=0$. This means $x_0^2+y_0^2=1$. Let $F(x,y,z)$ be the foot of the perpendicular from $P$ to the plane $2x+3y+z=6$. We need to find the locus of $F$.

Step 2: Relate the coordinates of the source point $P$ and the foot of the perpendicular $F$. The line segment $PF$ is perpendicular to the plane $2x+3y+z=6$. The direction ratios of the normal vector to this plane are $(2,3,1)$. Therefore, the line $PF$ has direction ratios $(2,3,1)$. Using the formula for the foot of the perpendicular, we can write the relationship between $P(x_0, y_0, 0)$ and $F(x,y,z)$ as: $\frac{x - x_0}{2} = \frac{y - y_0}{3} = \frac{z - 0}{1} = \lambda$ where $\lambda$ is a parameter.

Step 3: Express $x_0$ and $y_0$ in terms of $x, y, z,$ and $\lambda$. From the relationship in Step 2:

• $x - x_0 = 2\lambda \implies x_0 = x - 2\lambda$
• $y - y_0 = 3\lambda \implies y_0 = y - 3\lambda$
• $z = \lambda$

Step 4: Eliminate the parameter $\lambda$ using $z$. Since $z = \lambda$, we can substitute this into the expressions for $x_0$ and $y_0$:

• $x_0 = x - 2z$
• $y_0 = y - 3z$ This step is crucial as it connects the coordinates of the original point $P$ (which satisfies the circle equation) to the coordinates of the foot $F$ (whose locus we are finding).

Step 5: Use the equation of the circle for the source point $P$. We know that the point $P(x_0, y_0, 0)$ lies on the circle $x^2+y^2=1$. Therefore, its coordinates must satisfy this equation: $x_0^2 + y_0^2 = 1$ Substitute the expressions for $x_0$ and $y_0$ from Step 4 into this equation: $(x - 2z)^2 + (y - 3z)^2 = 1$ This equation now represents a relationship between the coordinates $(x,y,z)$ of the foot of the perpendicular.

Step 6: Use the equation of the plane for the foot of the perpendicular $F$. The foot of the perpendicular $F(x,y,z)$ lies on the plane $2x+3y+z=6$. This means its coordinates must satisfy the plane equation: $2x+3y+z=6$ From this, we can express $z$ in terms of $x$ and $y$: $z = 6 - 2x - 3y$ This step ensures that the point $(x,y,z)$ is indeed on the given plane.

Step 7: Substitute $z$ into the locus equation and simplify. Now, substitute the expression for $z$ from Step 6 into the equation derived in Step 5: $(x - 2(6 - 2x - 3y))^2 + (y - 3(6 - 2x - 3y))^2 = 1$ Simplify the terms inside the parentheses:

• First term: $x - 12 + 4x + 6y = (1+4)x + 6y - 12 = 5x + 6y - 12$
• Second term: $y - 18 + 6x + 9y = 6x + (1+9)y - 18 = 6x + 10y - 18$

So the equation becomes: $(5x + 6y - 12)^2 + (6x + 10y - 18)^2 = 1$ Notice that the second term $(6x + 10y - 18)$ has a common factor of 2. Factor it out: $(6x + 10y - 18) = 2(3x + 5y - 9)$ Squaring this term gives: $(6x + 10y - 18)^2 = (2(3x + 5y - 9))^2 = 4(3x + 5y - 9)^2$ Substitute this back into the locus equation: $(5x + 6y - 12)^2 + 4(3x + 5y - 9)^2 = 1$ This equation, along with $z = 6 - 2x - 3y$, defines the locus of the foot of the perpendicular.

Step 8: Match the derived equation with the given options. Our derived locus equation is: $(5x + 6y - 12)^2 + 4(3x + 5y - 9)^2 = 1$ And the $z$-coordinate is given by $z = 6 - 2x - 3y$. This matches option (B).

3. Common Mistakes & Tips

• Incorrect Normal Vector: Ensure the direction ratios of the line $PF$ are correctly taken as the coefficients of $x, y, z$ from the plane equation.
• Algebraic Errors: Be very careful with substitution and simplification, especially when dealing with multiple variables and parentheses. A small sign or coefficient error can lead to an incorrect option.
• Locus Definition: Remember that the final locus equation should only contain the coordinates of the point whose locus is being found (in this case, $x, y, z$) and eliminate any parameters or coordinates of the initial moving point.

4. Summary

To find the locus of the foot of the perpendicular from a point on a curve to a plane, we first express the coordinates of the point on the curve (source point) in terms of the coordinates of the foot of the perpendicular using the plane's normal vector. Then, we substitute these expressions into the equation of the original curve. Finally, we use the fact that the foot of the perpendicular lies on the plane to eliminate any remaining extra variables (like $z$) from the locus equation, resulting in an equation in $x$ and $y$. Following these steps, the derived locus is $(5x + 6y - 12)^2 + 4(3x + 5y - 9)^2 = 1$, with $z = 6 - 2x - 3y$.

The final answer is $\boxed{\text{(A)}}$.