If
\[
\left|
\begin{matrix}
x-3& 2& 1\\
3& x-2& 1\\
3& 2& x-1
\end{matrix}
\right|
=
(x-2)(px^{2}+qx+r)+8,
\]
then \(3p-2q+r=\)
Show Hint
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The Correct Option is B
Solution and Explanation
Step 1: Expand the determinant.
Let \[ D= \begin{vmatrix} x-3& 2& 1\\ 3& x-2& 1\\ 3& 2& x-1 \end{vmatrix}. \] Using expansion along the first row, \[ D=(x-3) \begin{vmatrix} x-2& 1\\ 2& x-1 \end{vmatrix} -2 \begin{vmatrix} 3& 1\\ 3& x-1 \end{vmatrix} + \begin{vmatrix} 3& x-2\\ 3& 2 \end{vmatrix}. \]
Step 2: Evaluate the minors.
\[ \begin{vmatrix} x-2& 1\\ 2& x-1 \end{vmatrix} =(x-2)(x-1)-2 =x^{2}-3x. \] \[ \begin{vmatrix} 3& 1\\ 3& x-1 \end{vmatrix} =3(x-1)-3 =3x-6. \] \[ \begin{vmatrix} 3& x-2\\ 3& 2 \end{vmatrix} =6-3(x-2) =12-3x. \]
Step 3: Obtain the polynomial.
\[ D=(x-3)(x^{2}-3x)-2(3x-6)+(12-3x). \] Simplifying, \[ D=x^{3}-6x^{2}+9x-6x+12+12-3x. \] \[ D=x^{3}-6x^{2}+24. \]
Step 4: Compare with the given expression.
Given \[ D=(x-2)(px^{2}+qx+r)+8. \] Therefore \[ x^{3}-6x^{2}+16 = (x-2)(px^{2}+qx+r). \] Dividing by \((x-2)\), \[ p=1,\qquad q=-4,\qquad r=-8. \]
Step 5: Find the required value.
\[ 3p-2q+r = 3(1)-2(-4)-8. \] \[ =3+8-8. \] \[ =3. \] Hence \[ \boxed{3p-2q+r=3}. \] Therefore the correct option is \[ \boxed{(B)}. \]
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