If the position vectors of the points A, B, C | vedclass

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(D) Given position vectors are:$A = \hat{i}+2 \hat{j}+3 \hat{k}$$B = 2 \hat{i}-\hat{j}+2 \hat{k}$$C = \frac{7}{4} \hat{i}+\frac{15}{4} \hat{j}+\frac{1

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$187$

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(D) Given position vectors are:$A = \hat{i}+2 \hat{j}+3 \hat{k}$$B = 2 \hat{i}-\hat{j}+2 \hat{k}$$C = \frac{7}{4} \hat{i}+\frac{15}{4} \hat{j}+\frac{15}{4} \hat{k}$$D = \frac{7}{3} \hat{i}+\frac{2}{3} \hat{j}+\frac{5+3a}{3} \hat{k}$Calculate vector $\vec{AC} = C - A = (\frac{7}{4}-1) \hat{i} + (\frac{15}{4}-2) \hat{j} + (\frac{15}{4}-3) \hat{k} = \frac{3}{4} \hat{i} + \frac{7}{4} \hat{j} + \frac{3}{4} \hat{k}$.Calculate vector $\vec{BD} = D - B = (\frac{7}{3}-2) \hat{i} + (\frac{2}{3}-(-1)) \hat{j} + (\frac{5+3a}{3}-2) \hat{k} = \frac{1}{3} \hat{i} + \frac{5}{3} \hat{j} + \frac{3a-1}{3} \hat{k}$.Given $|AC| = |BD|$, so $|AC|^2 = |BD|^2$:$(\frac{3}{4})^2 + (\frac{7}{4})^2 + (\frac{3}{4})^2 = (\frac{1}{3})^2 + (\frac{5}{3})^2 + (\frac{3a-1}{3})^2$$\frac{9+49+9}{16} = \frac{1+25+(3a-1)^2}{9}$$\frac{67}{16} = \frac{26+(3a-1)^2}{9}$$603 = 16(26 + (3a-1)^2)$$603 = 416 + 16(3a-1)^2$$16(3a-1)^2 = 603 - 416 = 187$.

If the position vectors of the points $A, B, C, D$ given by $\hat{i}+2 \hat{j}+3 \hat{k}, 2 \hat{i}-\hat{j}+2 \hat{k}$, $\frac{1}{4}(7 \hat{i}+15 \hat{j}+15 \hat{k})$ and $\frac{1}{3}[7 \hat{i}+2 \hat{j}+(5+3 a) \hat{k}]$ respectively are such that $|AC|=|BD|$, then $16(3a-1)^2=$

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