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Hint- Here, we will be finding the coordinates of the centre and the radius of the given sphere and then we will put the distance of the centre of the sphere from the tangential plane equal to the radius of the sphere.

Complete step-by-step answer:

Given equation of sphere is ${x^2} + {y^2} + {z^2} - 2y - 6z + 5 = 0{\text{ }} \to {\text{(1)}}$

As we know that the equation of any plane parallel to the given plane $ax + by + cz + d = 0$ is given by $ax + by + cz + d + \alpha = 0$

So, the equation of the plane parallel to the plane $2x + 2y - z = 0$ is given by

$2x + 2y - z + \alpha = 0{\text{ }} \to {\text{(2)}}$

Also we know that for a plane (having direction ratios as a, b and c) whose equation is given by $ax + by + cz + d = 0$ to be a tangent plane to any sphere ${\left( {x - {x_1}} \right)^2} + {\left( {y - {y_1}} \right)^2} + {\left( {z - {z_1}} \right)^2} = {R^2}{\text{ }} \to {\text{(3)}}$ where the centre of the sphere is ${\text{C}}\left( {{x_1},{y_1},{z_1}} \right)$ and radius as R, the perpendicular distance of the centre of the sphere ${\text{C}}\left( {{x_1},{y_1},{z_1}} \right)$ from the plane should be equal to the radius of the sphere.

Equation (1) of the sphere can be represented in the same form as the general equation of the sphere given by equation (3) with the help of completing the square method as shown under.

$

\Rightarrow {x^2} + {y^2} - 2y + {1^2} - {1^2} + {z^2} - 2 \times 3 \times z + {3^2} - {3^2} + 5 = 0 \\

\Rightarrow {x^2} + {\left( {y - 1} \right)^2} - 1 + {\left( {z - 3} \right)^2} - {3^2} + 5 = 0 \\

\Rightarrow {x^2} + {\left( {y - 1} \right)^2} + {\left( {z - 3} \right)^2} = 1 + 9 - 5 \\

\Rightarrow {\left( {x - 0} \right)^2} + {\left( {y - 1} \right)^2} + {\left( {z - 3} \right)^2} = 5 \\

\Rightarrow {x^2} + {\left( {y - 1} \right)^2} + {\left( {z - 3} \right)^2} = {\left( {\sqrt 5 } \right)^2}{\text{ }} \to {\text{(4)}} \\

$

By comparing equations (3) and (4), we get

${x_1} = 0,{y_1} = 1,{z_1} = 3$ and $R = \sqrt 5 $

The given sphere ${x^2} + {y^2} + {z^2} - 2y - 6z + 5 = 0$ have its centre at C(0,1,3) and radius as $R = \sqrt 5 $

Also, we know that the perpendicular distance of the point ${\text{P}}\left( {{x_1},{y_1},{z_1}} \right)$ from the plane $ax + by + cz + d = 0$ is given by

$d = \left| {\dfrac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|$

Since, it is given that the plane $2x + 2y - z + \alpha = 0$parallel to the given plane is a tangential plane to the given sphere which means that the perpendicular distance of centre C(0,1,3) from this plane will be equal to the radius of the sphere.

i.e., $

d = R \\

\Rightarrow \left| {\dfrac{{2{x_1} + 2{y_1} - {z_1} + \alpha }}{{\sqrt {{2^2} + {2^2} + {{\left( { - 1} \right)}^2}} }}} \right| = \sqrt 5 \\

$

Put the values of ${x_1} = 0,{y_1} = 1,{z_1} = 3$, we get

$

\Rightarrow \left| {\dfrac{{2 \times 0 + 2 \times 1 - 3 + \alpha }}{{\sqrt {4 + 4 + 1} }}} \right| = \sqrt 5 \\

\Rightarrow \left| {\dfrac{{2 - 3 + \alpha }}{{\sqrt 9 }}} \right| = \sqrt 5 \\

\Rightarrow \left| {\dfrac{{\alpha - 1}}{3}} \right| = \sqrt 5 \\

$

Either $

\dfrac{{\alpha - 1}}{3} = \sqrt 5 \\

\Rightarrow \alpha - 1 = 3\sqrt 5 \\

\Rightarrow \alpha = 1 + 3\sqrt 5 \\

$ or $

- \left( {\dfrac{{\alpha - 1}}{3}} \right) = \sqrt 5 \\

\Rightarrow \dfrac{{\alpha - 1}}{3} = - \sqrt 5 \\

\Rightarrow \alpha - 1 = - 3\sqrt 5 \\

\Rightarrow \alpha = 1 - 3\sqrt 5 \\

$

Put the values of $\alpha $ obtained in the equation (2), we get

When $\alpha = 1 + 3\sqrt 5 $,equation of the tangential plane is $2x + 2y - z + \left( {1 + 3\sqrt 5 } \right) = 0$

When $\alpha = 1 - 3\sqrt 5 $, equation of the tangential plane is $2x + 2y - z + \left( {1 - 3\sqrt 5 } \right) = 0$

So, the required equation of the tangential plane is $2x + 2y - z + \left( {1 \pm 3\sqrt 5 } \right) = 0$.

Hence, option C is correct.

Note- In this particular problem, the given plane $2x + 2y - z = 0$ has direction ratios as 2,2 and -1 which are the coefficients of x, y and z. It is important to note that any two planes if parallel should have same direction ratios. Also, there is a modulus present in the formula for perpendicular distance which results in two cases (one positive and other negative).

Complete step-by-step answer:

Given equation of sphere is ${x^2} + {y^2} + {z^2} - 2y - 6z + 5 = 0{\text{ }} \to {\text{(1)}}$

As we know that the equation of any plane parallel to the given plane $ax + by + cz + d = 0$ is given by $ax + by + cz + d + \alpha = 0$

So, the equation of the plane parallel to the plane $2x + 2y - z = 0$ is given by

$2x + 2y - z + \alpha = 0{\text{ }} \to {\text{(2)}}$

Also we know that for a plane (having direction ratios as a, b and c) whose equation is given by $ax + by + cz + d = 0$ to be a tangent plane to any sphere ${\left( {x - {x_1}} \right)^2} + {\left( {y - {y_1}} \right)^2} + {\left( {z - {z_1}} \right)^2} = {R^2}{\text{ }} \to {\text{(3)}}$ where the centre of the sphere is ${\text{C}}\left( {{x_1},{y_1},{z_1}} \right)$ and radius as R, the perpendicular distance of the centre of the sphere ${\text{C}}\left( {{x_1},{y_1},{z_1}} \right)$ from the plane should be equal to the radius of the sphere.

Equation (1) of the sphere can be represented in the same form as the general equation of the sphere given by equation (3) with the help of completing the square method as shown under.

$

\Rightarrow {x^2} + {y^2} - 2y + {1^2} - {1^2} + {z^2} - 2 \times 3 \times z + {3^2} - {3^2} + 5 = 0 \\

\Rightarrow {x^2} + {\left( {y - 1} \right)^2} - 1 + {\left( {z - 3} \right)^2} - {3^2} + 5 = 0 \\

\Rightarrow {x^2} + {\left( {y - 1} \right)^2} + {\left( {z - 3} \right)^2} = 1 + 9 - 5 \\

\Rightarrow {\left( {x - 0} \right)^2} + {\left( {y - 1} \right)^2} + {\left( {z - 3} \right)^2} = 5 \\

\Rightarrow {x^2} + {\left( {y - 1} \right)^2} + {\left( {z - 3} \right)^2} = {\left( {\sqrt 5 } \right)^2}{\text{ }} \to {\text{(4)}} \\

$

By comparing equations (3) and (4), we get

${x_1} = 0,{y_1} = 1,{z_1} = 3$ and $R = \sqrt 5 $

The given sphere ${x^2} + {y^2} + {z^2} - 2y - 6z + 5 = 0$ have its centre at C(0,1,3) and radius as $R = \sqrt 5 $

Also, we know that the perpendicular distance of the point ${\text{P}}\left( {{x_1},{y_1},{z_1}} \right)$ from the plane $ax + by + cz + d = 0$ is given by

$d = \left| {\dfrac{{a{x_1} + b{y_1} + c{z_1} + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|$

Since, it is given that the plane $2x + 2y - z + \alpha = 0$parallel to the given plane is a tangential plane to the given sphere which means that the perpendicular distance of centre C(0,1,3) from this plane will be equal to the radius of the sphere.

i.e., $

d = R \\

\Rightarrow \left| {\dfrac{{2{x_1} + 2{y_1} - {z_1} + \alpha }}{{\sqrt {{2^2} + {2^2} + {{\left( { - 1} \right)}^2}} }}} \right| = \sqrt 5 \\

$

Put the values of ${x_1} = 0,{y_1} = 1,{z_1} = 3$, we get

$

\Rightarrow \left| {\dfrac{{2 \times 0 + 2 \times 1 - 3 + \alpha }}{{\sqrt {4 + 4 + 1} }}} \right| = \sqrt 5 \\

\Rightarrow \left| {\dfrac{{2 - 3 + \alpha }}{{\sqrt 9 }}} \right| = \sqrt 5 \\

\Rightarrow \left| {\dfrac{{\alpha - 1}}{3}} \right| = \sqrt 5 \\

$

Either $

\dfrac{{\alpha - 1}}{3} = \sqrt 5 \\

\Rightarrow \alpha - 1 = 3\sqrt 5 \\

\Rightarrow \alpha = 1 + 3\sqrt 5 \\

$ or $

- \left( {\dfrac{{\alpha - 1}}{3}} \right) = \sqrt 5 \\

\Rightarrow \dfrac{{\alpha - 1}}{3} = - \sqrt 5 \\

\Rightarrow \alpha - 1 = - 3\sqrt 5 \\

\Rightarrow \alpha = 1 - 3\sqrt 5 \\

$

Put the values of $\alpha $ obtained in the equation (2), we get

When $\alpha = 1 + 3\sqrt 5 $,equation of the tangential plane is $2x + 2y - z + \left( {1 + 3\sqrt 5 } \right) = 0$

When $\alpha = 1 - 3\sqrt 5 $, equation of the tangential plane is $2x + 2y - z + \left( {1 - 3\sqrt 5 } \right) = 0$

So, the required equation of the tangential plane is $2x + 2y - z + \left( {1 \pm 3\sqrt 5 } \right) = 0$.

Hence, option C is correct.

Note- In this particular problem, the given plane $2x + 2y - z = 0$ has direction ratios as 2,2 and -1 which are the coefficients of x, y and z. It is important to note that any two planes if parallel should have same direction ratios. Also, there is a modulus present in the formula for perpendicular distance which results in two cases (one positive and other negative).

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