Question:

<p><strong>Step 1: Expand the determinant</strong></p><p>Expand along the second row:</p><p>$$\begin{vmatrix} ^{n-2}C_{r-2} & ^{n-2}C_{r-1} & ^{n-2}C_r \\ -3 & 1 & 1 \\ 2 & -1 & 0 \end{vmatrix} = -(-3)\begin{vmatrix} ^{n-2}C_{r-1} & ^{n-2}C_r \\ -1 & 0 \end{vmatrix} + 1\begin{vmatrix} ^{n-2}C_{r-2} & ^{n-2}C_r \\ 2 & 0 \end{vmatrix} - 1\begin{vmatrix} ^{n-2}C_{r-2} & ^{n-2}C_{r-1} \\ 2 & -1 \end{vmatrix}$$</p><p>$$= 3(0 + ^{n-2}C_r) + 1(0 - 2\cdot^{n-2}C_r) - 1(-^{n-2}C_{r-2} - 2\cdot^{n-2}C_{r-1})$$</p><p>$$= 3\cdot^{n-2}C_r - 2\cdot^{n-2}C_r + ^{n-2}C_{r-2} + 2\cdot^{n-2}C_{r-1}$$</p><p>$$= ^{n-2}C_{r-2} + 2\cdot^{n-2}C_{r-1} + ^{n-2}C_r$$</p><p><strong>Step 2: Apply Pascal's identity</strong></p><p>Using Pascal's identity: $^{n-2}C_{r-2} + ^{n-2}C_{r-1} = ^{n-1}C_{r-1}$ and $^{n-2}C_{r-1} + ^{n-2}C_r = ^{n-1}C_r$</p><p>$$^{n-2}C_{r-2} + 2\cdot^{n-2}C_{r-1} + ^{n-2}C_r = (^{n-2}C_{r-2} + ^{n-2}C_{r-1}) + (^{n-2}C_{r-1} + ^{n-2}C_r)$$</p><p>$$= ^{n-1}C_{r-1} + ^{n-1}C_r = ^nC_r$$</p><p><strong>Step 3: Evaluate the sum</strong></p><p>$$\sum_{r=2}^{n} (-2)^r \cdot ^nC_r = \sum_{r=0}^{n} (-2)^r\cdot^nC_r - ^nC_0 - 2\cdot(-2)^1\cdot^nC_1$$</p><p>Using the binomial theorem: $(1-2)^n = \sum_{r=0}^{n} (-2)^r\cdot^nC_r = (-1)^n$</p><p>$$(-1)^n - 1 - 2(-2)(n) = (-1)^n - 1 + 4n$$</p><p><strong>Step 4: Verify against options</strong></p><p>The result is $4n - 1 + (-1)^n$, which does not match options A, B, or C.</p><p>$$∴ \text{Answer: D}$$</p>