`a_(1), a_(2),a_(3) in R - {0} and a_(1)+ a_(2)cos2x+ a_(3)sin^(2)x=0` " for all " x in R then (a) vectors `veca=a_(1) hati+ a_(2) hatj + a_(3) hatk and vecb = 4hati + 2hatj + hatk` are perpendicular to each other (b)vectors `veca=a_(1) hati+ a_(2) hatj + a_(3) hatk and vecb = hati + hatj + 2hatk` are parallel to each each other (c)if vector `veca=a_(1) hati+ a_(2) hatj + a_(3) hatk` is of length `sqrt6` units, then on of the ordered trippplet `(a_(1), a_(2),a_(3)) = (1, -1,-2)` (d)if `2a_(1) + 3 a_(2) + 6 a_(3) = 26 , " then " |veca hati + a_(2) hatj + a_(3) hatk | is 2sqrt6`

Options

A. vectors `veca=a_(1) hati+ a_(2) hatj + a_(3) hatk and vecb = 4hati + 2hatj + hatk` are perpendicular to each other
B. vectors `veca=a_(1) hati+ a_(2) hatj + a_(3) hatk and vecb = hati + hatj + 2hatk` are parallel to each each other
C. if vector `veca=a_(1) hati+ a_(2) hatj + a_(3) hatk` is of length `sqrt6` units, then on of the ordered trippplet `(a_(1), a_(2),a_(3)) = (1, -1,-2)`
D. if `2a_(1) + 3 a_(2) + 6 a_(3) + 6a_(3) = 26 , " then " |veca hati + a_(2) hatj + a_(3) hatk | is 2sqrt6`

Explanation

To solve the problem, we start with the equation given:

\[ a_1 + a_2 \cos(2x) + a_3 \sin^2(x) = 0 \]

This equation must hold for all \( x \) in \( \mathbb{R} \).

### Step 1: Rewrite the equation using trigonometric identities
We can use the identity \( \cos(2x) = 1 - 2\sin^2(x) \) to rewrite the equation. Thus, we have:

\[ a_1 + a_2 (1 - 2\sin^2(x)) + a_3 \sin^2(x) = 0 \]

Expanding this gives:

\[ a_1 + a_2 - 2a_2 \sin^2(x) + a_3 \sin^2(x) = 0 \]

### Step 2: Collect terms involving \(\sin^2(x)\)
Now, we can group the terms:

\[ (a_1 + a_2) + (a_3 - 2a_2) \sin^2(x) = 0 \]

### Step 3: Set coefficients to zero
Since this equation must hold for all \( x \), the coefficients of the constant term and the coefficient of \(\sin^2(x)\) must both equal zero:

1. \( a_1 + a_2 = 0 \)
2. \( a_3 - 2a_2 = 0 \)

### Step 4: Solve the system of equations
From the first equation, we can express \( a_2 \) in terms of \( a_1 \):

\[ a_2 = -a_1 \]

Substituting \( a_2 \) into the second equation gives:

\[ a_3 - 2(-a_1) = 0 \implies a_3 + 2a_1 = 0 \implies a_3 = -2a_1 \]

Thus, we have:

\[ (a_1, a_2, a_3) = (a_1, -a_1, -2a_1) \]

### Step 5: Form the vector \(\vec{a}\)
Now we can express the vector \(\vec{a}\):

\[ \vec{a} = a_1 \hat{i} - a_1 \hat{j} - 2a_1 \hat{k} = a_1 (\hat{i} - \hat{j} - 2\hat{k}) \]

### Step 6: Check the options

#### Option (a): Check if vectors are perpendicular
Let \(\vec{b} = 4\hat{i} + 2\hat{j} + \hat{k}\).

To check if \(\vec{a}\) and \(\vec{b}\) are perpendicular, we compute the dot product:

\[ \vec{a} \cdot \vec{b} = a_1 (4 - 2 - 2) = 0 \]

Thus, \(\vec{a}\) and \(\vec{b}\) are perpendicular. **Option (a) is correct.**

#### Option (b): Check if vectors are parallel
Let \(\vec{b} = \hat{i} + \hat{j} + 2\hat{k}\).

For \(\vec{a}\) and \(\vec{b}\) to be parallel, there must exist a scalar \(\lambda\) such that:

\[ \vec{b} = \lambda \vec{a} \]

This does not hold for any \(\lambda\). **Option (b) is incorrect.**

#### Option (c): Check the length of \(\vec{a}\)
The length of \(\vec{a}\) is given by:

\[ |\vec{a}| = |a_1| \sqrt{1 + 1 + 4} = |a_1| \sqrt{6} \]

If \( |\vec{a}| = \sqrt{6} \), then:

\[ |a_1| \sqrt{6} = \sqrt{6} \implies |a_1| = 1 \implies a_1 = 1 \text{ or } -1 \]

Thus, one possible triplet is \( (1, -1, -2) \). **Option (c) is correct.**

#### Option (d): Check the equation
Given \( 2a_1 + 3a_2 + 6a_3 = 26 \):

Substituting \( a_2 = -a_1 \) and \( a_3 = -2a_1 \):

\[ 2a_1 + 3(-a_1) + 6(-2a_1) = 26 \implies 2a_1 - 3a_1 - 12a_1 = 26 \implies -13a_1 = 26 \implies a_1 = -2 \]

Now, substituting back, we find:

\[ |\vec{a}| = |-2| \sqrt{6} = 2\sqrt{6} \]

**Option (d) is correct.**

### Final Conclusion
The correct options are (a), (c), and (d).