Let $\vec{a}=4\overset{^}{i}-\overset{^}{j}+3\overset{^}{k},\vec{b}=10\overset{^}{i}+2\overset{^}{j}-\overset{^}{k}$ and a vector $\vec{c}$ be such that $2(\vec{a} \times \vec{b})+3(\vec{b} \times \vec{c})=\overrightarrow{0}$. If $\vec{a} \cdot \vec{c}=15$, then $\vec{c} \cdot(\hat{i}+\hat{j}-3 \hat{k})$ is equal to: [JEE Main 2026, 8 Apr (Shift 2)]

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YESPYQ · Accepted Answer

Correct answer: b) $-5$. $2(\vec{a} \times \vec{b})+3(\vec{b} \times \vec{c})=\overrightarrow{0}$ $(2 \vec{a}-3 \vec{c}) \times \vec{b}=\overrightarrow{0}$ $\vec{b} \|(2 \vec{a}-3 \vec{c})$ $2 \vec{a}-3 \vec{c}=\lambda \vec{b}$ $\overrightarrow{\mathrm{c}}=\frac{2 \overrightarrow{\mathrm{a}}-\lambda \overrightarrow{\mathrm{b}}}{3}$ $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=15$ $\left(\frac{2 \vec{a}-\lambda \vec{b}}{3}\right) \cdot \vec{a}=15$ $2(26)-\lambda(40-2-3)=45$ $\lambda=\frac{1}{5}$ $\Rightarrow \overrightarrow{\mathrm{c}}.(\hat{\mathrm{i}}+\hat{\mathrm{j}}-3 \hat{\mathrm{k}})=\frac{\left(2(4 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}})-\frac{1}{5}(10 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\mathrm{k})\right) \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}-3 \hat{\mathrm{k}})}{3}$ $=\frac{2(4-1-9)-\frac{1}{5}(10+2+3)}{3}=-5$

Let \(\vec{a}=4\overset{^}{i}-\overset{^}{j}+3\overset{^}{k},\vec{b}=10\overset{^}{i}+2\overset{^}{j}-\overset{^}{k}\) a…

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