Find the equation of the plane through the point (1, 2, 3) and parallel to the plane
\[x + 2y + 5z = 0.\]

a) \[(x - 1) + 2(y - 2) + 5(z - 3) = 0\]
b) \[x + 2y + 5z = 14\]
c) \[x + 2y + 5z = 6\]
d) None of these

Hint:
The normal vector to the given plane is also a normal vector to the parallel plane. We have a point and normal vector. So, we can use the Cartesian equation of the plane passing through the given point and containing the normal vector formula to get the answer.



Formula Used:The normal vector to a plane \[ax + by + cz + d = 0\]is given by \[\overrightarrow n = a\widehat i + b\widehat j + c\widehat k\].

The Cartesian equation of the plane passing through the given point and containing the normal vector is

\[a(x - {x_1}) + b(y - {y_1}) + c(z - {z_1}) = 0\]

Where the normal is vector \[\overrightarrow n = a\widehat i + b\widehat j + c\widehat k\] and \[({x_1},{y_1},{z_1})\]be the point on the plane.


Complete step by step solution:The normal vector to a plane \[x + 2y + 5z = 0\]is given by \[\overrightarrow n = \widehat i + 2\widehat j + 5\widehat k\].

Since the required plane is parallel to \[x + 2y + 5z = 0\], then the normal vector to the required plane is the same as the normal vector to \[x + 2y + 5z = 0\].

Hence, \[\overrightarrow n = \widehat i + 2\widehat j + 5\widehat k\] is the normal vector to the required plane.

The required plane passes through the point (1, 2, 3).

The Cartesian equation of the plane passing through the given point and containing the normal vector is

\[a(x - {x_1}) + b(y - {y_1}) + c(z - {z_1}) = 0\]

Here (a, b, c) = (1, 2, 5) and \[({x_1},{y_1},{z_1}) = (1,\,2,\,3).\]

\[1(x - 1) + 2(y - 2) + 5(z - 3) = 0\]

\[(x - 1) + 2(y - 2) + 5(z - 3) = 0\]



Option ‘A’ is correct



Note: This question can also be solved using the following method.

 The equation of the plane parallel to a given plane \[x + 2y + 5z = 0\]is of the form
\[x + 2y + 5z = k\] ---(1)

It passes through the point (1, 2, 3).

So equation (1) will satisfy the point (1, 2, 3).

\[ \Rightarrow x + 2y + 5z = k\]

 \[1 + 2(2) + 5(3) = k\]

  \[1 + 4 + 15 = k\]

             \[k = 20\]

Hence, the required equation is \[x + 2y + 5z = 20.\]

Option (b) and option (c) do not match the above answer.

Let us simplify the option (a).

\[(x - 1) + 2(y - 2) + 5(z - 3) = 0\]

          \[x - 1 + 2y - 4 + 5z - 15 = 0\]

                           \[x + 2y + 5z = 20\] which matches the above answer.

So, the correct choice is a.