Key Concepts and Formulas

1. Direction Cosines of a Vector: For a vector $\vec{v}$ making angles $\alpha, \beta, \gamma$ with the positive $x, y, z$ axes respectively, its direction cosines are $l = \cos \alpha$, $m = \cos \beta$, and $n = \cos \gamma$. These direction cosines satisfy the fundamental identity $l^2 + m^2 + n^2 = 1$. A vector in the direction of $\vec{v}$ can be represented by a scalar multiple of $l\hat{i} + m\hat{j} + n\hat{k}$.
2. Equation of a Plane: The equation of a plane passing through a point $P_0(x_0, y_0, z_0)$ and having a normal vector $\vec{N} = A\hat{i} + B\hat{j} + C\hat{k}$ is given by $A(x-x_0) + B(y-y_0) + C(z-z_0) = 0$.
3. Condition for Points on a Plane: If a plane passes through two points $P_1$ and $P_2$, then the vector connecting these two points, $\vec{P_1P_2}$, lies entirely within the plane. Consequently, $\vec{P_1P_2}$ must be perpendicular to the normal vector $\vec{N}$ of the plane. Mathematically, this means their dot product is zero: $\vec{N} \cdot \vec{P_1P_2} = 0$.

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Step-by-Step Solution

Step 1: Determine the Direction Cosines of Vector $\vec{v}$

1. Identify given angles: The problem states that vector $\vec{v}$ is inclined to the $x$-axis at $60^{\circ}$, to the $y$-axis at $45^{\circ}$, and to the $z$-axis at an acute angle.

   • Angle with $x$-axis ($\alpha$) $= 60^{\circ}$.
   • Angle with $y$-axis ($\beta$) $= 45^{\circ}$.
   • Angle with $z$-axis ($\gamma$) is acute, meaning $0^{\circ} < \gamma < 90^{\circ}$, so $\cos \gamma > 0$.

2. Calculate direction cosines $l$ and $m$:

   • $l = \cos \alpha = \cos 60^{\circ} = \frac{1}{2}$.
   • $m = \cos \beta = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$.

3. Calculate direction cosine $n$: We use the identity $l^2 + m^2 + n^2 = 1$.

   • $(\frac{1}{2})^2 + (\frac{1}{\sqrt{2}})^2 + n^2 = 1$
   • $\frac{1}{4} + \frac{1}{2} + n^2 = 1$
   • $\frac{3}{4} + n^2 = 1$
   • $n^2 = 1 - \frac{3}{4} = \frac{1}{4}$
   • Since $\gamma$ is an acute angle, $n = \cos \gamma$ must be positive. Therefore, $n = \sqrt{\frac{1}{4}} = \frac{1}{2}$.

4. Formulate the normal vector: The direction cosines of $\vec{v}$ are $(l, m, n) = (\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2})$. Since the plane is normal to $\vec{v}$, we can use $\vec{v}$ (or any scalar multiple of it) as the normal vector to the plane.

   • Let $\vec{N} = \frac{1}{2}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{2}\hat{k}$.
   • For simplicity in calculations, we can scale this normal vector by multiplying by $2\sqrt{2}$ (the least common multiple of the denominators) to get a proportional normal vector with integer/simpler coefficients: $\vec{N}' = 2\sqrt{2} \left( \frac{1}{2}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{2}\hat{k} \right) = \sqrt{2}\hat{i} + 2\hat{j} + \sqrt{2}\hat{k}$.
   • Alternatively, we can multiply the direction cosines $(1/2, 1/\sqrt{2}, 1/2)$ by 2 to get direction ratios $(1, \sqrt{2}, 1)$. This gives the normal vector $\vec{N}'' = \hat{i} + \sqrt{2}\hat{j} + \hat{k}$. This will be simpler for calculations.

Step 2: Determine the Equation of the Plane

1. Identify a point on the plane: The plane passes through the point $P_1(\sqrt{2}, -1, 1)$.
2. Use the normal vector and point to write the plane equation: Using $\vec{N}'' = \hat{i} + \sqrt{2}\hat{j} + \hat{k}$ as the normal vector and $P_1(\sqrt{2}, -1, 1)$ as a point on the plane, the equation of the plane is:
   • $A(x-x_1) + B(y-y_1) + C(z-z_1) = 0$
   • $1(x-\sqrt{2}) + \sqrt{2}(y-(-1)) + 1(z-1) = 0$
   • $1(x-\sqrt{2}) + \sqrt{2}(y+1) + 1(z-1) = 0$
   • $x - \sqrt{2} + \sqrt{2}y + \sqrt{2} + z - 1 = 0$
   • $x + \sqrt{2}y + z - 1 = 0$
   • $x + \sqrt{2}y + z = 1$

Step 3: Apply the condition for the second point

1. Substitute the second point into the plane equation: The plane also passes through the point $P_2(a, b, c)$. Since $(a, b, c)$ lies on the plane, it must satisfy the plane's equation.
   • $a + \sqrt{2}b + c = 1$

This establishes the relationship between $a, b,$ and $c$.

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Common Mistakes & Tips

• Forgetting $l^2+m^2+n^2=1$: This is the most crucial identity for direction cosines. Always use it to find the third direction cosine if two are known.
• Sign of direction cosines: For a vector in the first octant, all direction cosines ($l, m, n$) must be positive.
• Scalar multiples of normal vectors: Any non-zero scalar multiple of a normal vector is also a valid normal vector. Use this to simplify coefficients for easier calculations.
• Understanding "normal to a plane": A vector normal to a plane is perpendicular to every vector lying within that plane. This is key to using the dot product condition.

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Summary

First, we determined the direction cosines of the vector $\vec{v}$ using the given angles and the identity $l^2+m^2+n^2=1$. This yielded the direction cosines $(\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2})$. We then used these direction cosines to form a normal vector for the plane. With one point on the plane $(\sqrt{2},-1,1)$ and the normal vector, we derived the equation of the plane as $x+\sqrt{2}y+z=1$. Finally, by substituting the second point $(a,b,c)$ into the plane's equation, we found the required relationship: $a+\sqrt{2}b+c=1$.

The final answer is $\boxed{a+\sqrt{2} b+c=1}$, which corresponds to option (D).